Question

Find an equation in standard form for the hyperbola with vertices at (0,+-6) and asymptote at Y= +- 3/7x

Answer 1

`(y^2)/(36)-(x^2)/(196)=1`

Explanation:

From the question we are told that:

Vertices of hyperbola at
`(0,\pm6)`

Asymptotes at
`Y= \pm 3/7x`

Generally the expression for Vertices of hyperbola is mathematically given by

`(0,\pm a)`

Therefore

`(0,\pm a)=(0,\pm6)`

Comparing variables

`a=6`

Generally the expression for Asymptotes of hyperbola is mathematically given by

`y=\pm (a)/(b)x`

Therefore

`\pm (a)/(b)x= \pm (3)/(7)x`

Therefore

`(6)/(b)= (3)/(7)`

`b=(42)/(3)`

`b=14`

Generally the equation for Hyperbola of hyperbola is mathematically given by

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

`(y^2)/(6^2)-(x^2)/(14^2)=1`

`(y^2)/(36)-(x^2)/(196)=1`