Question

Find an equation in standard form for the hyperbola with vertices at (0,plus or minus 6) and asymptotes at y= (plus or minus 3/5)x

Answer 1

`((y)^2)/(36)-((x)^2)/(100)=1`

Explanation:

Given:

Vertices of Hyperbola : (0 ± 6) or (0,6) and (0,-6)

and asymptotes at y= (±3/5)x 0r y= 3/5 x and y=-3/5 x

The vertices are of vertical hyperbola. The equation used will be:

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

The Center of hyperbola (h,k) =(0,0)

The Distance from vertices to center is a and a = 6 (given)

For equation we have value of h,k and a and need to find value of b

we know,

y= k ± a/b (x-h)

Values of h and k are zero

y= 0 ± a/b (x-0)

y= (± a/b ) x

We are given asymtotes at y= (± 3/5)x which is equal to y= (± a/b ) x

as a = 6 then b= 10 i.e The simplified form of 6/10 is 3/5 so value of b=10

Putting values of a,b,h and k in equation we get,

`((y-0)^2)/((6)^2)-((x-0)^2)/((10)^2)=1\\\\((y)^2)/(36)-((x)^2)/(100)=1`