Question

(1 point) Find an equation of the ellipse with foci at (−5,9) and (−5,−10) and whose major axis has length 22. Express your answer in the form P(x,y)=0, where P(x,y) is a polynomial in x and y such that the coefficient of x 2

is 121 . Answer: =0

Answer 1

To find the equation of the ellipse with foci at (-5,9) and (-5,-10) and a major axis length of 22, use the formula
`(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1` The center of the ellipse is (-5,-5) and the semiminor axis is 11. The value of a is the distance from the center to one of the foci. We find equation to be
`121(x + 5)^2 - 96(y + 5)^2 = 121 * 96`

Step-by-step explanation:

To find the equation of the ellipse with foci at (-5,9) and (-5,-10) and a major axis length of 22, we can use the formula for the distance between two points in an ellipse:

`(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1`

Where (h,k) is the center of the ellipse and a and b are the lengths of the semimajor and semiminor axes, respectively. In this case, since the foci are in the vertical line y = -5, the center of the ellipse is (-5,-5) and the semiminor axis b is half the length of the major axis, so b = 22/2 = 11. Now we need to find the value of a, which is the distance from the center to one of the foci:

`a^2 = c^2 - b^2`

`a^2 = (-10 - (-5))^2 - 11^2`

`a^2 = 5^2 - 11^2`

`a^2 = -96`

Since a^2 is negative, we know that the equation of the ellipse is:

`(x + 5)^2 / (-96) + (y + 5)^2 / 121 = 1`

This can be rewritten as:

`121(x + 5)^2 - 96(y + 5)^2 = 121 * 96`