In the standard (x, y) coordinate plane, an ellipse is defined by the equation 7x^2 + 16y^… — ACT

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

Answer: 13

Answer

The larger of the two

y

-coordinates of the points on the parabola is 13.

By completing the square on the general ellipse equation, we get

\frac{(x-3)^2}{16} + \frac{(y-1)^2}{7} = 1

. The center is

(3, 1)

and the focal distance is

c = \sqrt{16-7} = 3

, meaning the foci are at

(0, 1)

and

(6, 1)

. The parabola has its vertex at

(0, 1)

and focus at

(6, 1)

, which means it opens to the right with

p = 6

. Its equation is

(y - 1)^2 = 24x

. Substituting

x = 6

yields

(y - 1)^2 = 144

, so

y - 1 = \pm 12

. The two possible

y

-coordinates are

13

and

-11

, of which

13

is the larger value.

Step-by-Step Solution

Complete the square for the given ellipse equation to rewrite it in standard form.

\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.

Find the focal distance

c

and calculate the coordinates of the foci.

Focal distance

c = 3

; Foci at

(0, 1)

and

(6, 1)

For an ellipse, the distance

c

from the center

(h, k)

to the foci is

\sqrt{a^2 - b^2}

. Since the major axis is horizontal, the foci are located at

(h \pm c, k)

Use the foci coordinates to identify the vertex and focus of the parabola.

Vertex:

(0, 1)

; Focus:

(6, 1)

The problem defines the parabola's vertex as the ellipse focus with the smaller

x

-coordinate, and the parabola's focus as the ellipse focus with the larger

x

-coordinate.

Determine the equation of the parabola using its vertex and focus.

(y - 1)^2 = 24x

The parabola is horizontal and opens to the right with focal distance

p = 6

. The standard form is

(y - k)^2 = 4p(x - h)

Substitute

x = 6

into the parabola equation and solve for the larger

y

-value.

y = 13

Substituting

x = 6

yields

(y - 1)^2 = 144

, which gives

y = 1 + 12 = 13

y = 1 - 12 = -11

. The larger value is 13.

Key Concept

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

Alternative Method

Once the equation

(y - 1)^2 = 24x

is established, recognize that at

x = 6

(which is the

x

-coordinate of the focus), the points on the parabola form the endpoints of the latus rectum. The length of the latus rectum is

4p = 24

, so the points lie at distance

2p = 12

vertically above and below the focus

(6, 1)

. Thus, the

y

-coordinates are

1 \pm 12

, immediately yielding the larger coordinate as 13.

Estimated Time:3m 0s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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