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Zorluk: OrtaConic Sections

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?

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
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Konular

Conic Sections

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