Question

Drag each sign and value to the correct location on the image. Each sign and value can be used more than once, but not all signs and values will be used. The vertices of an ellipse are at (-5, -2) and (-5, 14), and the point (0, 6) lies on the ellipse. Drag the missing terms and signs to their correct places in the standard form of the equation of this ellipse.

Answer 1

The complete equation of the ellipse is $\frac{(x + 5)^2}{5^2} + \frac{(y - 6)^2}{8^2} = 1$

How to complete the vertex equation?

The complete question is in the attachment

The given parameters are:

• Vertex = (-5,-2) and (-5,14)
• Point = (-0,6)

The vertex is represented as (h, k ± a).

So, we have:

h = -5

k + a = 14

k - a = -2

Add the last two equations

2k = 12

Divide by 2

k = 6

Substitute k = 6 in k + a = 14

6 + a = 14

Solve for a

a = 8

The ellipse equation is represented as:

$\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$

So, we have:

$\frac{(x + 5)^2}{b^2} + \frac{(y - 6)^2}{8^2} = 1$

The ellipse passes through the point (0,6).

So, we have:

$\frac{(0 + 5)^2}{b^2} + \frac{(6 - 6)^2}{8^2} = 1$

This gives

$\frac{5^2}{b^2} + \frac{0}{8^2} = 1$

Evaluate the quotient

$\frac{5^2}{b^2} = 1$

Cross multiply

$b^2 = 5^2$

Take the square root of both sides

b = 5

Substitute b = 5 in $\frac{(x + 5)^2}{b^2} + \frac{(y - 6)^2}{8^2} = 1$

$\frac{(x + 5)^2}{5^2} + \frac{(y - 6)^2}{8^2} = 1$

Hence, the complete equation of the ellipse is $\frac{(x + 5)^2}{5^2} + \frac{(y - 6)^2}{8^2} = 1$

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