Question

Given the hyperbola (y2/16) – (x2/9) = 1, find the equations for its asymptotes.​

Answer 1

ANSWER:

Since your hyperbola has the "y" term listed first, it opens "up and down.'

It's also centered at (0,0)

So the form is: y²/b² - x^²/a² = 1

In your case,

b² = 16

so, b = √(16) = 4.

And a² = 9

so, a = √(9) = 3.

So, the equations for the asymptotes are y = +(b/a)x and y = -(b/a)x

So we have, in your case,

y = +(4/3)x and y = - (4/3)x.

Answer 2

As this equation is already in the correct format, put x and y equal to zero and you will see that when x = 0, y² = 16 which gives the y intercepts.

If you put y = 0 , you get x² = − 9 which doesn't give any real roots. The graph will therefore be a hyperbola in the shape of a V and an upside down V (as opposed to being on their sides like ><).

You can use the formula:

y = ±(b/a)x

to determine the equations of the asymptotes where

a = √9, b = √16

and you will get

y = ± (4/3)x.