Examkin

Conic Sections

19 soru

Tüm türler19Çoktan seçmeli10Sayısal9
Soru 1Soru →

A circle in the standard (x,y)(x, y) coordinate plane has its center at (5,−2)(5, -2) and a radius of 77 units. Which of the following is an equation of this circle?

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Cevap: x2+y2−10x+4y=20x^2 + y^2 - 10x + 4y = 20

Cevap

The equation x2+y2−10x+4y=20x^2 + y^2 - 10x + 4y = 20
The standard form equation of a circle is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center and rr is the radius. Substituting h=5h = 5, k=−2k = -2, and r=7r = 7 yields (x−5)2+(y+2)2=49(x - 5)^2 + (y + 2)^2 = 49. Expanding both binomials gives x2−10x+25+y2+4y+4=49x^2 - 10x + 25 + y^2 + 4y + 4 = 49. Combining the constant terms on the left side gives x2+y2−10x+4y+29=49x^2 + y^2 - 10x + 4y + 29 = 49. Subtracting 2929 from both sides results in the equation x2+y2−10x+4y=20x^2 + y^2 - 10x + 4y = 20.

Adım Adım Çözüm

1
Write down the standard equation of a circle.
(x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center and rr is the radius.
This is the fundamental formula relating the geometric properties of a circle to its algebraic representation.
2
Substitute the given values into the standard equation.
(x−5)2+(y−(−2))2=72(x - 5)^2 + (y - (-2))^2 = 7^2, which simplifies to (x−5)2+(y+2)2=49(x - 5)^2 + (y + 2)^2 = 49.
Substituting the center (5,−2)(5, -2) and radius 77 into the standard equation sets up the expression for expansion.
3
Expand the squared binomials.
x2−10x+25+y2+4y+4=49x^2 - 10x + 25 + y^2 + 4y + 4 = 49.
Expanding (x−5)2(x - 5)^2 into x2−10x+25x^2 - 10x + 25 and (y+2)2(y + 2)^2 into y2+4y+4y^2 + 4y + 4 allows conversion to the general form.
4
Simplify and rearrange the equation to match the form of the options.
x2+y2−10x+4y+29=49x^2 + y^2 - 10x + 4y + 29 = 49, which simplifies to x2+y2−10x+4y=20x^2 + y^2 - 10x + 4y = 20.
Combining the constants and subtracting 2929 from both sides of the equation yields the final simplified general form.

Anahtar Kavram

Equation of a Circle in General Form
Tahmini Süre:1m 0s
Soru 2Soru →

An ellipse in the standard coordinate plane is defined by the equation 9x2+25y2−36x+50y−164=09x^2 + 25y^2 - 36x + 50y - 164 = 0. What is the distance between the two foci of this ellipse?

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Cevap: 8

Cevap

8
The correct answer is 8. Rearranging the given equation 9x2+25y2−36x+50y−164=09x^2 + 25y^2 - 36x + 50y - 164 = 0 by completing the square gives the standard form (x−2)225+(y+1)29=1\frac{(x-2)^2}{25} + \frac{(y+1)^2}{9} = 1. In this standard horizontal ellipse equation, the semi-major axis squared is a2=25a^2 = 25 and the semi-minor axis squared is b2=9b^2 = 9. The focal distance cc from the center to each focus is found using the relation c2=a2−b2c^2 = a^2 - b^2, which yields c=25−9=4c = \sqrt{25 - 9} = 4. Since the distance between the two foci is 2c2c, the final distance is 2(4)=82(4) = 8.

Adım Adım Çözüm

1
Group the xx and yy terms and move the constant to the right-hand side.
(9x2−36x)+(25y2+50y)=164(9x^2 - 36x) + (25y^2 + 50y) = 164
Grouping like variables allows us to factor out coefficients before completing the square.
2
Factor out the leading coefficients of the quadratic terms.
9(x2−4x)+25(y2+2y)=1649(x^2 - 4x) + 25(y^2 + 2y) = 164
Completing the square requires the quadratic terms inside the parentheses to have a coefficient of 1.
3
Complete the square for both variables by adding the square of half the linear coefficients inside the parentheses, and balance the equation by adding the distributed values to the right side.
9(x2−4x+4)+25(y2+2y+1)=164+9(4)+25(1)9(x^2 - 4x + 4) + 25(y^2 + 2y + 1) = 164 + 9(4) + 25(1) which simplifies to 9(x−2)2+25(y+1)2=2259(x-2)^2 + 25(y+1)^2 = 225.
This rewrites the quadratic expressions into perfect square binomials.
4
Divide both sides of the equation by 225 to write the equation in standard form.
(x−2)225+(y+1)29=1\frac{(x-2)^2}{25} + \frac{(y+1)^2}{9} = 1
The standard form of an ellipse equation is equal to 1.
5
Identify the values of a2a^2 and b2b^2 to calculate the distance cc from the center to each focus.
a2=25a^2 = 25 and b2=9b^2 = 9. Using c2=a2−b2c^2 = a^2 - b^2, we get c2=25−9=16c^2 = 25 - 9 = 16, so c=4c = 4.
For a horizontal ellipse, the larger denominator is a2a^2 and the smaller is b2b^2, and the focal distance satisfies c2=a2−b2c^2 = a^2 - b^2.
6
Multiply the focal distance from the center by 2 to find the total distance between the two foci.
Distance=2c=2(4)=8\text{Distance} = 2c = 2(4) = 8.
The distance between the two foci is the length of the segment connecting them, which is centered at (2,−1)(2, -1) and extends cc units in both horizontal directions.

Anahtar Kavram

Rewriting an ellipse equation in standard form to determine its key geometric features including its foci.
Tahmini Süre:1m 30s
Soru 3Soru →

A parabola in the standard (x,y)(x, y) coordinate plane is defined by the equation y=112x2−2x+13y = \frac{1}{12}x^2 - 2x + 13. What is the yy-coordinate of the focus of this parabola?

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Cevap: 4

Cevap

The y-coordinate of the focus is 4.
The standard form of the parabola is y−1=112(x−12)2y - 1 = \frac{1}{12}(x - 12)^2. Comparing this to y−k=14p(x−h)2y - k = \frac{1}{4p}(x - h)^2 gives the vertex (h,k)=(12,1)(h, k) = (12, 1) and 4p=12  ⟹  p=34p = 12 \implies p = 3. Since the parabola opens upward, the focus is at (12,1+3)=(12,4)(12, 1 + 3) = (12, 4), making the yy-coordinate 4.

Adım Adım Çözüm

1
Complete the square to rewrite the equation in standard vertex form.
y=112(x−12)2+1y = \frac{1}{12}(x - 12)^2 + 1
Rewriting the general quadratic equation into standard form allows us to directly identify the vertex and focal parameters.
2
Equate the coefficients to find the focal distance pp and vertex (h,k)(h, k).
Vertex (h,k)=(12,1)(h, k) = (12, 1) and p=3p = 3
The standard vertex form of a vertical parabola is y−k=14p(x−h)2y - k = \frac{1}{4p}(x - h)^2. Setting 14p=112\frac{1}{4p} = \frac{1}{12} gives p=3p = 3.
3
Calculate the focus coordinates (h,k+p)(h, k + p).
Focus =(12,4)= (12, 4), so the yy-coordinate is 44
For an upward-opening parabola, the focus is located pp units directly above the vertex.

Anahtar Kavram

Rewriting a quadratic equation into standard vertex form to find the properties of a parabola, including its vertex and focus.
Soru 4Soru →

A parabola in the standard (x,y)(x, y) coordinate plane is defined by the equation y2=12xy^2 = 12x. What is the xx-coordinate of the focus of this parabola?

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Cevap: 3

Cevap

The xx-coordinate of the focus is 3.
The equation y2=12xy^2 = 12x represents a parabola with vertex at the origin opening to the right. The standard form for such a parabola is y2=4pxy^2 = 4px, where the focus is located at (p,0)(p, 0). By setting 4p=124p = 12, we find p=3p = 3. Thus, the focus is (3,0)(3, 0), and its xx-coordinate is 3.

Adım Adım Çözüm

1
Identify the standard form of the parabola's equation.
The equation y2=12xy^2 = 12x fits the standard form of a horizontal parabola with its vertex at the origin, y2=4pxy^2 = 4px.
This allows us to relate the given equation to the coordinate of the focus, which is located at (p,0)(p, 0).
2
Solve for the parameter pp.
4p=124p = 12, which gives p=3p = 3.
By equating the coefficients of xx from the given equation and the standard form, we can find the value of pp.
3
Determine the focus coordinate.
The focus is at (3,0)(3, 0), so the xx-coordinate is 3.
The focus of a parabola of the form y2=4pxy^2 = 4px has coordinates (p,0)(p, 0).

Anahtar Kavram

Focus of a Parabola
Soru 5Soru →

A hyperbola in the standard coordinate plane is represented by the equation 9x2−16y2−36x−32y−124=09x^2 - 16y^2 - 36x - 32y - 124 = 0. Which of the following equations represents one of the asymptotes of this hyperbola?

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Cevap: y=34x−52y = \frac{3}{4}x - \frac{5}{2}

Cevap

y=34x−52y = \frac{3}{4}x - \frac{5}{2}
The correct equation is found by rewriting the hyperbola equation in standard form (x−2)216−(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1 through completing the square. This indicates a horizontal hyperbola centered at (2,−1)(2, -1) with a=4a = 4 and b=3b = 3. The asymptotes are given by y−k=±ba(x−h)y - k = \pm \frac{b}{a}(x - h). Substituting the values gives y+1=±34(x−2)y + 1 = \pm \frac{3}{4}(x - 2). Simplifying the positive case results in the correct equation.

Adım Adım Çözüm

1
Group the xx-terms and yy-terms and factor out the coefficients of the squared terms.
9(x2−4x)−16(y2+2y)=1249(x^2 - 4x) - 16(y^2 + 2y) = 124
Grouping prepares the algebraic expression for completing the square for both variables.
2
Complete the square for both the xx and yy expressions by adding the balanced constants to the right side.
9(x−2)2−16(y+1)2=1449(x - 2)^2 - 16(y + 1)^2 = 144
Adding 9(4)=369(4) = 36 and subtracting 16(1)=1616(1) = 16 to the right side balances the equation after completing the square.
3
Divide both sides by 144144 to express the equation in the standard form of a hyperbola.
(x−2)216−(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1
The standard form equation (x−h)2a2−(y−k)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 reveals the center (h,k)(h, k) and the semi-axes values aa and bb.
4
Identify the key parameters of the hyperbola and state the general formula for its asymptotes.
Center (h,k)=(2,−1)(h, k) = (2, -1), a=4a = 4, b=3b = 3. The asymptotes are given by y−k=±ba(x−h)y - k = \pm \frac{b}{a}(x - h), which simplifies to y+1=±34(x−2)y + 1 = \pm \frac{3}{4}(x - 2).
For a horizontal hyperbola, the rise-over-run slope of the asymptotes is governed by the ratio ba\frac{b}{a}.
5
Simplify the positive slope case to find the matching slope-intercept equation.
y=34x−52y = \frac{3}{4}x - \frac{5}{2}
Distributing the slope gives y+1=34x−32y + 1 = \frac{3}{4}x - \frac{3}{2}, and subtracting 11 from both sides yields the final equation.

Anahtar Kavram

Rewriting a hyperbola equation using completing the square to find its asymptotes.

Alternatif Yöntem

Instead of completing the square entirely, find the center of the hyperbola by taking partial derivatives. The derivative with respect to xx is 18x−36=0  ⟹  x=218x - 36 = 0 \implies x = 2. The derivative with respect to yy is −32y−32=0  ⟹  y=−1-32y - 32 = 0 \implies y = -1. Thus, the center is (2,−1)(2, -1). The slope of the asymptotes can be found from the ratio of the square roots of the coefficients of the quadratic terms: m=±916=±34m = \pm \sqrt{\frac{9}{16}} = \pm \frac{3}{4}. Using the point-slope form with the center (2,−1)(2, -1) and slope 34\frac{3}{4} gives y+1=34(x−2)y + 1 = \frac{3}{4}(x - 2), which simplifies directly to y=34x−52y = \frac{3}{4}x - \frac{5}{2}.
Tahmini Süre:2m 30s
Soru 6Soru →

In the standard (x,y)(x, y) coordinate plane, a hyperbola is defined by the equation 9x2−16y2−54x−64y−127=09x^2 - 16y^2 - 54x - 64y - 127 = 0. What is the shortest distance from the focus of the hyperbola with the larger xx-coordinate to the asymptote with the positive slope?

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Cevap: 3

Cevap

3
The correct answer is 3. Completing the square for 9x2−16y2−54x−64y−127=09x^2 - 16y^2 - 54x - 64y - 127 = 0 yields the standard form equation (x−3)216−(y+2)29=1\frac{(x-3)^2}{16} - \frac{(y+2)^2}{9} = 1. The focus with the larger xx-coordinate is at (8,−2)(8, -2) and the asymptote with the positive slope is 3x−4y−17=03x - 4y - 17 = 0. Applying the point-to-line distance formula yields a distance of 3.

Adım Adım Çözüm

1
Group the xx-terms and yy-terms and move the constant to the right side of the equation.
9(x2−6x)−16(y2+4y)=1279(x^2 - 6x) - 16(y^2 + 4y) = 127
Grouping the terms allows us to complete the square for the xx and yy variables separately.
2
Complete the square for both the xx and yy expressions, adjusting the right side of the equation by adding the weighted constants.
9(x2−6x+9)−16(y2+4y+4)=127+9(9)−16(4)  ⟹  9(x−3)2−16(y+2)2=1449(x^2 - 6x + 9) - 16(y^2 + 4y + 4) = 127 + 9(9) - 16(4) \implies 9(x-3)^2 - 16(y+2)^2 = 144
Completing the square allows us to write the quadratic expressions as perfect squares to put the equation in standard form.
3
Divide both sides of the equation by 144 to obtain the standard form of the hyperbola.
(x−3)216−(y+2)29=1\frac{(x-3)^2}{16} - \frac{(y+2)^2}{9} = 1
The standard form of a horizontal hyperbola is (x−h)2a2−(y−k)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1, which reveals the center (h,k)(h, k) and the semi-axes aa and bb.
4
Identify the key parameters of the hyperbola: center, aa, bb, and calculate the focal distance cc.
Center is (3,−2)(3, -2), a=4a = 4, b=3b = 3, and c=a2+b2=16+9=5c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = 5.
These parameters are required to find the coordinates of the focus and the equation of the asymptote.
5
Find the coordinates of the focus with the larger xx-coordinate.
Focus is (3+5,−2)=(8,−2)(3 + 5, -2) = (8, -2).
For a horizontal hyperbola, the foci are located at (h±c,k)(h \pm c, k). The focus with the larger xx-coordinate is at (h+c,k)(h+c, k).
6
Determine the equation of the asymptote with the positive slope.
The asymptote equation is y+2=34(x−3)  ⟹  3x−4y−17=0y + 2 = \frac{3}{4}(x - 3) \implies 3x - 4y - 17 = 0.
The asymptotes of a horizontal hyperbola are given by y−k=±ba(x−h)y - k = \pm \frac{b}{a}(x - h). The one with the positive slope uses +ba+\frac{b}{a}.
7
Use the point-to-line distance formula d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} to find the distance from the focus (8,−2)(8, -2) to the asymptote line 3x−4y−17=03x - 4y - 17 = 0.
d=∣3(8)−4(−2)−17∣32+(−4)2=∣24+8−17∣25=155=3d = \frac{|3(8) - 4(-2) - 17|}{\sqrt{3^2 + (-4)^2}} = \frac{|24 + 8 - 17|}{\sqrt{25}} = \frac{15}{5} = 3.
Calculating this gives the shortest distance from the focus to the asymptote.

Anahtar Kavram

Rewriting the general equation of a hyperbola into standard form by completing the square, identifying its center, foci, and asymptotes, and applying the distance formula from a point to a line.
Soru 7Soru →

A parabola in the standard (x,y)(x, y) coordinate plane is defined by the equation x2−6x−8y+25=0x^2 - 6x - 8y + 25 = 0. What is the distance, in coordinate units, between the focus and the directrix of this parabola?

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Cevap: 4

Cevap

The distance between the focus and the directrix of the parabola is 4.
By completing the square on the equation x2−6x−8y+25=0x^2 - 6x - 8y + 25 = 0, we get (x−3)2=8(y−2)(x-3)^2 = 8(y-2). Since the coefficient of the linear factor is 88, we set 4p=84p = 8, which yields p=2p = 2. The distance from the focus to the directrix is 2p=2(2)=42p = 2(2) = 4.

Adım Adım Çözüm

1
Isolate the terms containing xx on one side of the equation.
x2−6x=8y−25x^2 - 6x = 8y - 25
To set up the equation for completing the square on the xx terms.
2
Complete the square for the quadratic expression in xx by adding 99 to both sides.
x2−6x+9=8y−16  ⟹  (x−3)2=8y−16x^2 - 6x + 9 = 8y - 16 \implies (x-3)^2 = 8y - 16
Adding (−6/2)2=9( -6/2 )^2 = 9 creates a perfect square trinomial on the left side.
3
Factor out the coefficient of yy on the right side to write the equation in standard form.
(x−3)2=8(y−2)(x-3)^2 = 8(y-2)
This matches the standard form equation (x−h)2=4p(y−k)(x-h)^2 = 4p(y-k) for a vertical parabola.
4
Determine the value of the focal parameter pp from the standard form.
4p=8  ⟹  p=24p = 8 \implies p = 2
Comparing the standard form coefficient 4p4p with the value 88 gives p=2p = 2.
5
Calculate the total distance between the focus and the directrix.
2p=2(2)=42p = 2(2) = 4
The vertex is situated halfway between the focus and the directrix, making the distance between them 2p2p.

Anahtar Kavram

Finding the geometric properties of a parabola by completing the square to convert its general equation to standard form.
Soru 8Soru →

An ellipse in the standard (x,y)(x, y) coordinate plane is defined by the equation x2+4y2+6x−8y+9=0x^2 + 4y^2 + 6x - 8y + 9 = 0. What is the length of the major axis of this ellipse?

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Cevap: 4

Cevap

4
To find the length of the major axis, rewrite the general equation of the ellipse in standard form by completing the square. Grouping the terms yields (x2+6x)+4(y2−2y)=−9(x^2 + 6x) + 4(y^2 - 2y) = -9. Completing the square for both variables gives (x+3)2−9+4[(y−1)2−1]=−9(x+3)^2 - 9 + 4[(y-1)^2 - 1] = -9, which simplifies to (x+3)2+4(y−1)2=4(x+3)^2 + 4(y-1)^2 = 4. Dividing both sides by 4 gives the standard form (x+3)24+(y−1)21=1\frac{(x+3)^2}{4} + \frac{(y-1)^2}{1} = 1. In this form, the horizontal axis is the major axis because the denominator under the xx-term (a2=4a^2 = 4) is larger than the denominator under the yy-term (b2=1b^2 = 1). Since a2=4a^2 = 4, the semi-major axis is a=2a = 2. Therefore, the total length of the major axis is 2a=2(2)=42a = 2(2) = 4.

Adım Adım Çözüm

1
Group the xx-terms and yy-terms and move the constant to the right side of the equation.
(x2+6x)+(4y2−8y)=−9(x^2 + 6x) + (4y^2 - 8y) = -9
Grouping like terms allows completing the square for each variable independently.
2
Factor out the coefficient of y2y^2 from the yy-terms.
(x2+6x)+4(y2−2y)=−9(x^2 + 6x) + 4(y^2 - 2y) = -9
Before completing the square, the leading coefficient of the squared terms inside the parentheses must be 1.
3
Complete the square for both the xx and yy expressions by adding and subtracting the square of half of the linear coefficients.
((x+3)2−9)+4((y−1)2−1)=−9((x+3)^2 - 9) + 4((y-1)^2 - 1) = -9
This rewrites the quadratic expressions into perfect square trinomial form.
4
Distribute the coefficients and simplify the constant terms.
(x+3)2−9+4(y−1)2−4=−9  ⟹  (x+3)2+4(y−1)2−13=−9  ⟹  (x+3)2+4(y−1)2=4(x+3)^2 - 9 + 4(y-1)^2 - 4 = -9 \implies (x+3)^2 + 4(y-1)^2 - 13 = -9 \implies (x+3)^2 + 4(y-1)^2 = 4
Isolating the squared terms on one side helps convert the equation to the standard form of an ellipse.
5
Divide both sides of the equation by 4 to set the right side equal to 1.
(x+3)24+(y−1)21=1\frac{(x+3)^2}{4} + \frac{(y-1)^2}{1} = 1
The standard form of a horizontal ellipse is (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1.
6
Identify the values of a2a^2 and b2b^2 and calculate the length of the major axis.
a2=4  ⟹  a=2a^2 = 4 \implies a = 2. The major axis length is 2a=2(2)=42a = 2(2) = 4.
The length of the major axis is twice the length of the semi-major axis (aa).

Anahtar Kavram

Rewriting the general equation of an ellipse into standard form by completing the square to find its key features, such as the length of the major axis.

Alternatif Yöntem

Another way to find the length of the major axis is to find the vertices of the ellipse by finding the maximum and minimum x-values where the equation has real solutions for y, though completing the square is the standard and most direct method.
Tahmini Süre:1m 30s
Soru 9Soru →

The equation of a parabola is given by (x−4)2=12(y+1)(x - 4)^2 = 12(y + 1). What is the yy-coordinate of the focus of this parabola?

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Cevap: 2

Cevap

The correct answer is 2.
The equation (x−4)2=12(y+1)(x - 4)^2 = 12(y + 1) is a parabola with a vertical axis of symmetry, vertex at (4,−1)(4, -1), and focal length p=3p = 3. The focus is located pp units above the vertex, yielding a yy-coordinate of −1+3=2-1 + 3 = 2.

Adım Adım Çözüm

1
Identify the standard form of the parabola's equation.
The equation (x−4)2=12(y+1)(x - 4)^2 = 12(y + 1) matches (x−h)2=4p(y−k)(x - h)^2 = 4p(y - k).
This form allows us to find the vertex and the focal distance pp directly.
2
Determine the vertex and focal distance pp.
The vertex is (4,−1)(4, -1) and p=3p = 3 since 4p=124p = 12.
Matching the given equation terms to the standard form reveals these properties.
3
Find the coordinates of the focus.
The focus is at (4,2)(4, 2).
The focus is located pp units vertically above the vertex for a parabola opening upward.

Anahtar Kavram

Focus of a Parabola
Soru 10Soru →

In the standard (x,y)(x, y) coordinate plane, an ellipse is defined by the equation 7x2+16y2−42x−32y−33=07x^2 + 16y^2 - 42x - 32y - 33 = 0. A parabola has its vertex at the focus of the ellipse with the smaller xx-coordinate, and its focus at the focus of the ellipse with the larger xx-coordinate. What is the larger of the two yy-coordinates of the points on the parabola that have an xx-coordinate of 6?

Cevabı ve açıklamayı göster

Cevap: 13

Cevap

The larger of the two yy-coordinates of the points on the parabola is 13.
By completing the square on the general ellipse equation, we get (x−3)216+(y−1)27=1\frac{(x-3)^2}{16} + \frac{(y-1)^2}{7} = 1. The center is (3,1)(3, 1) and the focal distance is c=16−7=3c = \sqrt{16-7} = 3, meaning the foci are at (0,1)(0, 1) and (6,1)(6, 1). The parabola has its vertex at (0,1)(0, 1) and focus at (6,1)(6, 1), which means it opens to the right with p=6p = 6. Its equation is (y−1)2=24x(y - 1)^2 = 24x. Substituting x=6x = 6 yields (y−1)2=144(y - 1)^2 = 144, so y−1=±12y - 1 = \pm 12. The two possible yy-coordinates are 1313 and −11-11, of which 1313 is the larger value.

Adım Adım Çözüm

1
Complete the square for the given ellipse equation to rewrite it in standard form.
(x−3)216+(y−1)27=1\frac{(x-3)^2}{16} + \frac{(y-1)^2}{7} = 1
Converting the equation to standard form is necessary to determine the center and semi-axis lengths of the ellipse.
2
Find the focal distance cc and calculate the coordinates of the foci.
Focal distance c=3c = 3; Foci at (0,1)(0, 1) and (6,1)(6, 1)
For an ellipse, the distance cc from the center (h,k)(h, k) to the foci is a2−b2\sqrt{a^2 - b^2}. Since the major axis is horizontal, the foci are located at (h±c,k)(h \pm c, k).
3
Use the foci coordinates to identify the vertex and focus of the parabola.
Vertex: (0,1)(0, 1); Focus: (6,1)(6, 1)
The problem defines the parabola's vertex as the ellipse focus with the smaller xx-coordinate, and the parabola's focus as the ellipse focus with the larger xx-coordinate.
4
Determine the equation of the parabola using its vertex and focus.
(y−1)2=24x(y - 1)^2 = 24x
The parabola is horizontal and opens to the right with focal distance p=6p = 6. The standard form is (y−k)2=4p(x−h)(y - k)^2 = 4p(x - h).
5
Substitute x=6x = 6 into the parabola equation and solve for the larger yy-value.
y=13y = 13
Substituting x=6x = 6 yields (y−1)2=144(y - 1)^2 = 144, which gives y=1+12=13y = 1 + 12 = 13 or y=1−12=−11y = 1 - 12 = -11. The larger value is 13.

Anahtar Kavram

Determining the equations and key features (foci, vertices, focal parameters) of ellipses and parabolas by rewriting equations into standard forms.

Alternatif Yöntem

Once the equation (y−1)2=24x(y - 1)^2 = 24x is established, recognize that at x=6x = 6 (which is the xx-coordinate of the focus), the points on the parabola form the endpoints of the latus rectum. The length of the latus rectum is 4p=244p = 24, so the points lie at distance 2p=122p = 12 vertically above and below the focus (6,1)(6, 1). Thus, the yy-coordinates are 1±121 \pm 12, immediately yielding the larger coordinate as 13.
Tahmini Süre:3m 0s
Soru 11Soru →

In the standard (x,y)(x, y) coordinate plane, an ellipse is centered at the origin (0,0)(0, 0) and has vertices at (−5,0)(-5, 0) and (5,0)(5, 0). If the distance between the two foci of the ellipse is 88, what is the length of the minor axis of the ellipse?

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Cevap: 6

Cevap

The correct answer is 6.
The correct answer is 6 because the ellipse has a horizontal major axis with a=5a = 5 and focal distance c=4c = 4. Using the relationship c2=a2−b2c^2 = a^2 - b^2, we solve for the semi-minor axis bb to get b=3b = 3. The total length of the minor axis is 2b=62b = 6.

Adım Adım Çözüm

1
Determine the semi-major axis length aa.
a=5a = 5
The vertices are at (±5,0)(\pm 5, 0), which are 55 units from the center (0,0)(0, 0) along the major axis.
2
Determine the distance from the center to each focus cc.
c=4c = 4
The distance between the two foci is 2c=82c = 8, so the distance from the center to a focus is c=4c = 4.
3
Find the semi-minor axis length bb.
b=3b = 3
Using the relation c2=a2−b2c^2 = a^2 - b^2 for ellipses, we get 42=52−b2  ⟹  b2=9  ⟹  b=34^2 = 5^2 - b^2 \implies b^2 = 9 \implies b = 3.
4
Calculate the full length of the minor axis.
66
The length of the minor axis is 2b=2(3)=62b = 2(3) = 6.

Anahtar Kavram

The relationship between the semi-major axis, semi-minor axis, and focal distance of an ellipse.
Tahmini Süre:1m 30s
Soru 12Soru →

An ellipse in the standard (x,y)(x, y) coordinate plane is defined by the equation 25x2+9y2−100x+54y−44=025x^2 + 9y^2 - 100x + 54y - 44 = 0. Which of the following points is a focus of this ellipse?

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Cevap: (2,1)(2, 1)

Cevap

(2,1)(2, 1)
The correct answer is the point (2,1)(2, 1) because rewriting the general equation of the ellipse in standard form gives (x−2)29+(y+3)225=1\frac{(x-2)^2}{9} + \frac{(y+3)^2}{25} = 1. The center is (2,−3)(2, -3) and the major axis is vertical with a focal distance of c=25−9=4c = \sqrt{25 - 9} = 4. Adding this distance to the yy-coordinate of the center yields the focus (2,−3+4)=(2,1)(2, -3 + 4) = (2, 1).

Adım Adım Çözüm

1
Group the xx and yy terms and factor out the coefficients.
25(x2−4x)+9(y2+6y)=4425(x^2 - 4x) + 9(y^2 + 6y) = 44
This prepares the algebraic equation for completing the square.
2
Complete the square for both the xx and yy groups.
25(x−2)2+9(y+3)2=22525(x-2)^2 + 9(y+3)^2 = 225
Adding 25×4=10025 \times 4 = 100 and 9×9=819 \times 9 = 81 to both sides maintains equality while converting the quadratic expressions into perfect square trinomials.
3
Divide both sides by 225225 to write the equation in standard form.
(x−2)29+(y+3)225=1\frac{(x-2)^2}{9} + \frac{(y+3)^2}{25} = 1
The standard form of an ellipse equation is (x−h)2b2+(y−k)2a2=1\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 (for a vertical major axis), which directly reveals the center (h,k)(h, k) and the axis parameters.
4
Identify the center, axis lengths, and calculate the focal distance cc.
Center is (2,−3)(2, -3), a2=25a^2 = 25, b2=9b^2 = 9. Thus, c=25−9=4c = \sqrt{25 - 9} = 4.
The focal distance cc for an ellipse is determined by the relation c=a2−b2c = \sqrt{a^2 - b^2}.
5
Determine the coordinates of the foci.
Foci are (2,−3±4)(2, -3 \pm 4), which simplifies to (2,1)(2, 1) and (2,−7)(2, -7).
Since a2=25a^2 = 25 is under the yy-term, the ellipse is vertically oriented, meaning the foci lie on the vertical line passing through the center.

Anahtar Kavram

Rewriting the general equation of an ellipse to standard form and finding its foci.
Tahmini Süre:2m 30s
Soru 13Soru →

A hyperbola in the standard (x,y)(x, y) coordinate plane is defined by the equation:

16x2−9y2−32x+36y−164=016x^2 - 9y^2 - 32x + 36y - 164 = 0

What are the coordinates of the foci of this hyperbola?

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Cevap: (6,2)(6, 2) and (−4,2)(-4, 2)

Cevap

The foci of the hyperbola are (6,2)(6, 2) and (−4,2)(-4, 2).
The correct answer is the set of coordinates (6,2)(6, 2) and (−4,2)(-4, 2). After rewriting the hyperbola's equation in standard form by completing the square, we obtain (x−1)29−(y−2)216=1\frac{(x - 1)^2}{9} - \frac{(y - 2)^2}{16} = 1. The center of the hyperbola is (1,2)(1, 2). Since the x2x^2 term is positive, it has a horizontal transverse axis. The distance from the center to each focus, cc, satisfies c2=a2+b2=9+16=25c^2 = a^2 + b^2 = 9 + 16 = 25, so c=5c = 5. Adding and subtracting this focal distance from the xx-coordinate of the center yields the foci at (1+5,2)=(6,2)(1 + 5, 2) = (6, 2) and (1−5,2)=(−4,2)(1 - 5, 2) = (-4, 2).

Adım Adım Çözüm

1
Group the xx-terms and yy-terms and move the constant to the right side of the equation.
16(x2−2x)−9(y2−4y)=16416(x^2 - 2x) - 9(y^2 - 4y) = 164
Grouping the terms allows us to prepare for completing the square for both variables.
2
Complete the square for x2−2xx^2 - 2x by adding 11 inside the parentheses, and for y2−4yy^2 - 4y by adding 44 inside the parentheses. Add the corresponding balanced quantities to the right side: 16(1)=1616(1) = 16 and −9(4)=−36-9(4) = -36.
16(x2−2x+1)−9(y2−4y+4)=164+16−3616(x^2 - 2x + 1) - 9(y^2 - 4y + 4) = 164 + 16 - 36
16(x−1)2−9(y−2)2=14416(x - 1)^2 - 9(y - 2)^2 = 144
This rewrites the quadratic expressions into perfect square binomials.
3
Divide both sides of the equation by 144144 to express it in standard form.
(x−1)29−(y−2)216=1\frac{(x - 1)^2}{9} - \frac{(y - 2)^2}{16} = 1
The standard form of a horizontal hyperbola is (x−h)2a2−(y−k)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1, which lets us identify the center, aa, and bb directly.
4
Identify the center (h,k)(h, k), a2a^2, and b2b^2, then calculate the focal distance cc using the relation c2=a2+b2c^2 = a^2 + b^2.
Center is (1,2)(1, 2). a2=9a^2 = 9 and b2=16b^2 = 16.
c2=9+16=25  ⟹  c=5c^2 = 9 + 16 = 25 \implies c = 5.
Foci are located at a distance of cc from the center along the transverse axis.
5
Determine the coordinates of the foci by shifting the xx-coordinate of the center by ±c\pm c since the transverse axis is horizontal.
Foci coordinates are (1±5,2)(1 \pm 5, 2), which gives (6,2)(6, 2) and (−4,2)(-4, 2).
Adding and subtracting cc from the center's xx-coordinate gives the locations of the two foci.

Anahtar Kavram

Rewriting a hyperbola equation in standard form by completing the square and finding its foci.
Soru 14Soru →

A hyperbola in the standard (x,y)(x, y) coordinate plane is defined by the equation 9x2−16y2−36x−32y−124=09x^2 - 16y^2 - 36x - 32y - 124 = 0. One of the foci of this hyperbola is located at the point (f,−1)(f, -1), where f>0f > 0. What is the value of ff?

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Cevap: 7

Cevap

7
Completing the square transforms the equation into the standard form of a horizontal hyperbola, (x−2)216−(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1, which has its center at (2,−1)(2, -1) with a2=16a^2 = 16 and b2=9b^2 = 9. The distance to the foci is c=a2+b2=16+9=5c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = 5. Since the transverse axis is horizontal, the foci are located at (2±5,−1)(2 \pm 5, -1), which are (−3,−1)(-3, -1) and (7,−1)(7, -1). Given the constraint that f>0f > 0, the positive x-coordinate of the focus is 7.

Adım Adım Çözüm

1
Group the terms and prepare to complete the square.
9(x2−4x)−16(y2+2y)=1249(x^2 - 4x) - 16(y^2 + 2y) = 124
Grouping the variables helps isolate the quadratic expressions for completing the square.
2
Complete the square for both the xx and yy terms.
9(x−2)2−16(y+1)2=1449(x - 2)^2 - 16(y + 1)^2 = 144
To complete the square for x2−4xx^2 - 4x, add 4 inside the first parentheses, adding 9×4=369 \times 4 = 36 to the right side. To complete the square for y2+2yy^2 + 2y, add 1 inside the second parentheses, which subtracts 16×1=1616 \times 1 = 16 from the right side because of the leading negative coefficient. This leaves the right side as 124+36−16=144124 + 36 - 16 = 144.
3
Divide both sides of the equation by the constant to find the standard form.
(x−2)216−(y+1)29=1\frac{(x - 2)^2}{16} - \frac{(y + 1)^2}{9} = 1
Dividing by 144 puts the equation in the standard horizontal hyperbola form: (x−h)2a2−(y−k)2b2=1\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1.
4
Find the distance cc from the center to the foci.
c=5c = 5
For a hyperbola, the relationship between the semi-axes and the focal distance is c=a2+b2c = \sqrt{a^2 + b^2}. Substituting a2=16a^2 = 16 and b2=9b^2 = 9 gives c=16+9=5c = \sqrt{16 + 9} = 5.
5
Determine the coordinates of the foci and extract the value of ff.
f=7f = 7
The center of the hyperbola is (h,k)=(2,−1)(h, k) = (2, -1). The foci are located at (h±c,k)=(2±5,−1)(h \pm c, k) = (2 \pm 5, -1), which corresponds to the points (−3,−1)(-3, -1) and (7,−1)(7, -1). Since the problem states f>0f > 0, the target focus must be (7,−1)(7, -1), meaning f=7f = 7.

Anahtar Kavram

Converting a general hyperbola equation into standard form to calculate focal points
Soru 15Soru →

A hyperbola is defined by the equation 9x2−4y2−36x−8y−4=09x^2 - 4y^2 - 36x - 8y - 4 = 0 in the standard (x,y)(x, y) coordinate plane. What is the slope of the asymptote of this hyperbola that has a positive slope?

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Cevap: 1.5

Cevap

The positive slope of the asymptotes is 1.5.
By converting the general form of the hyperbola equation into standard form, we determine that it is a horizontal hyperbola with a=2a = 2 and b=3b = 3. The slopes of the asymptotes for a horizontal hyperbola are ±ba\pm \frac{b}{a}, making the positive slope equal to 32=1.5\frac{3}{2} = 1.5.

Adım Adım Çözüm

1
Group x-terms and y-terms, and move the constant to the other side.
9(x2−4x)−4(y2+2y)=49(x^2 - 4x) - 4(y^2 + 2y) = 4
This prepares the equation for completing the square.
2
Complete the square for the quadratic expressions in x and y.
9(x−2)2−4(y+1)2=369(x - 2)^2 - 4(y + 1)^2 = 36
Completing the square yields 9[(x−2)2−4]−4[(y+1)2−1]=4  ⟹  9(x−2)2−36−4(y+1)2+4=49[(x-2)^2 - 4] - 4[(y+1)^2 - 1] = 4 \implies 9(x-2)^2 - 36 - 4(y+1)^2 + 4 = 4.
3
Divide both sides by 36 to format the equation in standard hyperbola form.
(x−2)24−(y+1)29=1\frac{(x-2)^2}{4} - \frac{(y+1)^2}{9} = 1
The standard form of a horizontal hyperbola centered at (h,k)(h, k) is (x−h)2a2−(y−k)2b2=1\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1.
4
Identify the values of a and b from the denominators.
a=2a = 2 and b=3b = 3
Since a2=4a^2 = 4 and b2=9b^2 = 9, taking the square roots gives a=2a = 2 and b=3b = 3.
5
Determine the positive slope of the asymptotes using the formula for a horizontal hyperbola.
m=ba=1.5m = \frac{b}{a} = 1.5
The asymptotes for a horizontal hyperbola are given by y−k=±ba(x−h)y - k = \pm \frac{b}{a}(x - h), so the positive slope is ba=32=1.5\frac{b}{a} = \frac{3}{2} = 1.5.

Anahtar Kavram

Rewriting a hyperbola equation from general form to standard form to find asymptote equations.
Tahmini Süre:1m 30s
Soru 16Soru →

A circle in the standard (x,y)(x, y) coordinate plane has center (3,−4)(3, -4) and passes through the point (6,0)(6, 0). Which of the following is an equation of this circle?

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Cevap: x2+y2−6x+8y=0x^2 + y^2 - 6x + 8y = 0

Cevap

x2+y2−6x+8y=0x^2 + y^2 - 6x + 8y = 0
The standard equation of a circle is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center and rr is the radius. Given the center is (3,−4)(3, -4), the equation becomes (x−3)2+(y+4)2=r2(x - 3)^2 + (y + 4)^2 = r^2. Since the circle passes through (6,0)(6, 0), we substitute these coordinates to find r2r^2: (6−3)2+(0+4)2=32+42=9+16=25(6 - 3)^2 + (0 + 4)^2 = 3^2 + 4^2 = 9 + 16 = 25. The equation is therefore (x−3)2+(y+4)2=25(x - 3)^2 + (y + 4)^2 = 25. Expanding this equation gives x2−6x+9+y2+8y+16=25x^2 - 6x + 9 + y^2 + 8y + 16 = 25. Combining constant terms yields x2+y2−6x+8y+25=25x^2 + y^2 - 6x + 8y + 25 = 25. Subtracting 25 from both sides results in the general form equation x2+y2−6x+8y=0x^2 + y^2 - 6x + 8y = 0.

Adım Adım Çözüm

1
Calculate the radius squared of the circle using the distance formula between the center (3,−4)(3, -4) and the point on the circle (6,0)(6, 0).
r2=(6−3)2+(0−(−4))2=32+42=9+16=25r^2 = (6 - 3)^2 + (0 - (-4))^2 = 3^2 + 4^2 = 9 + 16 = 25
The distance between the center and any point on the circle is equal to the radius of the circle.
2
Write the standard form of the circle's equation using the center (h,k)=(3,−4)(h, k) = (3, -4) and the radius squared r2=25r^2 = 25.
(x−3)2+(y+4)2=25(x - 3)^2 + (y + 4)^2 = 25
The standard equation of a circle is (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2.
3
Expand the squared binomials in the standard equation to convert it into general form.
x2−6x+9+y2+8y+16=25x^2 - 6x + 9 + y^2 + 8y + 16 = 25
Expanding allows us to combine like terms and match the general form expressions in the options.
4
Simplify the expanded equation by combining constant terms and setting the equation to zero.
x2+y2−6x+8y+25=25⇒x2+y2−6x+8y=0x^2 + y^2 - 6x + 8y + 25 = 25 \Rightarrow x^2 + y^2 - 6x + 8y = 0
Subtracting 25 from both sides yields the final simplified general form equation of the circle.

Anahtar Kavram

Deriving and expanding the equation of a circle from its center and a point.
Soru 17Soru →

An ellipse in the standard (x,y)(x, y) coordinate plane is defined by the equation (x−4)2169+(y+3)2144=1\frac{(x-4)^2}{169} + \frac{(y+3)^2}{144} = 1. What is the distance between the two foci of this ellipse?

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Cevap: 10

Cevap

The distance between the two foci of the ellipse is 10.
By comparing the given equation to the standard form of an ellipse, we find a2=169a^2 = 169 and b2=144b^2 = 144. The distance from the center to each focus, cc, is given by c=a2−b2=169−144=25=5c = \sqrt{a^2 - b^2} = \sqrt{169 - 144} = \sqrt{25} = 5. The total distance between the two foci is 2c=2(5)=102c = 2(5) = 10.

Adım Adım Çözüm

1
Identify a2a^2 and b2b^2 from the given equation of the ellipse.
a2=169a^2 = 169 and b2=144b^2 = 144
The standard equation of a horizontal ellipse is (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, where a2a^2 is the larger denominator.
2
Calculate the value of cc, the distance from the center to a focus.
c=5c = 5
For an ellipse, the focal distance cc is related to the semi-major axis aa and semi-minor axis bb by the equation c2=a2−b2c^2 = a^2 - b^2.
3
Calculate the distance between the two foci, which is 2c2c.
10
The distance between the two foci of an ellipse is twice the distance from the center to each focus (2c2c).

Anahtar Kavram

Focal distance of an ellipse
Tahmini Süre:1m 15s
Soru 18Soru →

A circle in the standard (x,y)(x, y) coordinate plane is defined by the equation x2+y2−4x+6y−12=0x^2 + y^2 - 4x + 6y - 12 = 0. What is the radius of this circle?

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Cevap: 55

Cevap

The radius of the circle is 5.
The correct answer is 5. Grouping the xx-terms and yy-terms of the equation x2+y2−4x+6y−12=0x^2 + y^2 - 4x + 6y - 12 = 0 gives (x2−4x)+(y2+6y)=12(x^2 - 4x) + (y^2 + 6y) = 12. Completing the square requires adding (−42)2=4(\frac{-4}{2})^2 = 4 and (62)2=9(\frac{6}{2})^2 = 9 to both sides, yielding (x2−4x+4)+(y2+6y+9)=12+4+9(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9, which simplifies to (x−2)2+(y+3)2=25(x-2)^2 + (y+3)^2 = 25. In the standard circle equation form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2, r2=25r^2 = 25, meaning the radius rr is 25=5\sqrt{25} = 5.

Adım Adım Çözüm

1
Group the xx-terms and yy-terms, and move the constant term to the right side of the equation.
(x2−4x)+(y2+6y)=12(x^2 - 4x) + (y^2 + 6y) = 12
This prepares the quadratic expression to be written in standard circle form by completing the square.
2
Find the constant values needed to complete the square for both the xx and yy variables. Add these constants to both sides of the equation.
(x2−4x+4)+(y2+6y+9)=12+4+9(x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9
The constant for xx is (−42)2=4(\frac{-4}{2})^2 = 4, and the constant for yy is (62)2=9(\frac{6}{2})^2 = 9. Adding them to both sides maintains equality.
3
Factor the perfect square trinomials on the left side and simplify the right side.
(x−2)2+(y+3)2=25(x-2)^2 + (y+3)^2 = 25
This puts the equation in the standard form (x−h)2+(y−k)2=r2(x-h)^2 + (y-k)^2 = r^2.
4
Identify r2r^2 from the standard form and solve for the radius rr.
r=25=5r = \sqrt{25} = 5
Taking the square root of the constant on the right side yields the radius of the circle.

Anahtar Kavram

Rewriting a circle's equation from general form to standard form by completing the square to find its radius.

Alternatif Yöntem

For a circle given in the general form x2+y2+Dx+Ey+F=0x^2 + y^2 + Dx + Ey + F = 0, the radius can be calculated directly using the formula r=12D2+E2−4Fr = \frac{1}{2}\sqrt{D^2 + E^2 - 4F}. Substituting the coefficients D=−4D = -4, E=6E = 6, and F=−12F = -12 gives: r=12(−4)2+62−4(−12)=1216+36+48=12100=12(10)=5r = \frac{1}{2}\sqrt{(-4)^2 + 6^2 - 4(-12)} = \frac{1}{2}\sqrt{16 + 36 + 48} = \frac{1}{2}\sqrt{100} = \frac{1}{2}(10) = 5.
Tahmini Süre:1m 30s
Soru 19Soru →

An ellipse in the standard (x,y)(x, y) coordinate plane is defined by the equation

(x−2)29+(y+5)216=1\frac{(x-2)^2}{9} + \frac{(y+5)^2}{16} = 1

What are the coordinates of the center of this ellipse?

Cevabı ve açıklamayı göster

Cevap: (2,−5)(2, -5)

Cevap

The center of the ellipse is (2,−5)(2, -5).
The standard form of an ellipse equation is (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, where the center of the ellipse is at the coordinate point (h,k)(h, k). Comparing the given equation (x−2)29+(y+5)216=1\frac{(x-2)^2}{9} + \frac{(y+5)^2}{16} = 1 to the standard form reveals that h=2h = 2 and k=−5k = -5. Therefore, the center coordinates are (2,−5)(2, -5).

Adım Adım Çözüm

1
Recall the standard form equation of an ellipse with horizontal/vertical axes.
The standard form is (x−h)2a2+(y−k)2b2=1\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1, where (h,k)(h, k) represents the coordinates of the center.
This establishes the framework to extract the center coordinate values.
2
Compare the terms in the given equation to the standard form.
Matching (x−h)2(x-h)^2 with (x−2)2(x-2)^2 gives h=2h = 2. Matching (y−k)2(y-k)^2 with (y+5)2=(y−(−5))2(y+5)^2 = (y-(-5))^2 gives k=−5k = -5.
Comparing terms identifies the offsets hh and kk that determine the center.
3
Write the center coordinate pair (h,k)(h, k).
The center is (2,−5)(2, -5).
Combining the values of hh and kk yields the final coordinates.

Anahtar Kavram

Identifying the center of an ellipse from its standard form equation
Tahmini Süre:45s