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

Answer:

$\frac{(y)^2}{36}-\frac{(x)^2}{100}=1$

Step-by-step 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:

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

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,

$\frac{(y-0)^2}{(6)^2}-\frac{(x-0)^2}{(10)^2}=1\\\\\frac{(y)^2}{36}-\frac{(x)^2}{100}=1$