Question

I need help on two questions!
1. Write the equation of an ellipse with vertices (10, 0) and (-10, 0) and co-vertices (0, 2) and (0, -2).

second question in below.

Answer 1

1) Answer: $\frac{x^{2}}{100} +\frac{y^{2}}{4} = 1$

Explanation:

The equation of an ellipse is: $\frac{(x-h)^{2}}{a^{2}} +\frac{(y-k)^{2}}{b^{2}} = 1$ ; where (h, k) is the center, "a" is the x-radius, and "b" is the y-radius.

               Center                        Radius

x-axis:   (10 + -10)/2 = 0            10 - 0 = 10

•       h = 0                             a = 10

y-axis:   (2 + -2)/2  = 0               2 - 0 = 2

•       k = 0                             b = 2

Now, input the values into the equation:

$\frac{(x-0)^{2}}{10^{2}} +\frac{(y-0)^{2}}{2^{2}} = 1$

$\frac{x^{2}}{100} +\frac{y^{2}}{4} = 1$

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2) Answer: $\frac{x^{2}}{1} +\frac{(y+2)^{2}}{4} = 1$

Explanation:

Vertices are: (0, 1) and (0, -5) ------> x-values are the same, y = 1, -5

Covertices are: (-1, -2) and (1, -2) ----> y-values are the same, x = -1, 1

               Center                        Radius

x-axis:   (-1 + 1)/2 = 0                  1 - 0 = 1

•       h = 0                             a = 1

y-axis:   (1 + -5)/2  = -2               1 - (-2) = 3

•       k = -2                             b = 3

Now, input the values into the equation:

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

$\frac{(x-0)^{2}}{1^{2}} +\frac{(y-(-2))^{2}}{3^{2}} = 1$

$\frac{x^{2}}{1} +\frac{(y+2)^{2}}{9} = 1$