Question

Write the equation of the conic section with the given properties:

Answer 1

Answer:

$\frac{x^2}{64} + \frac{y^2}{25} =1$

Step-by-step explanation:

An ellipse with vertices (-8, 0) and (8, 0)

Distance between two vertices = 2a

Distance between (-8,0) and (8,0) = 16

2a= 16

so a= 8

Vertex is (h+a,k)

we know a=8, so vertex is (h+8,k)

Now compare (h+8,k) with vertex (8,0) and find out h and k

h+8 =8, h=0

k =0  

a minor axis of length 10.

Length of minor axis = 2b

2b = 10

so b = 5

General formula for the equation of horizontal ellipse is

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

a= 8 , b=5 , h=0,k=0. equation becomes

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

$\frac{x^2}{64} + \frac{y^2}{25} =1$