Question

What are the equations for the asymptotes of this hyperbola?

Answer 1

Answer: Option D

`y=(6)/(11)x`

and

`y=-(6)/(11)x`

Explanation:

We know that the general equation of a hyperbola with transverse vertical axis has the form:

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

Where the point (h, k) are the coordinates of the center of the ellipse

2b is the length of the transverse horizontal axis

2a is the length of the conjugate axis

The asymptotes are lines to which the hyperbola graphic approaches but never touches.

The asymptotes have the following equation

`y -y_1=\±m(x-x_1)\\\\`

Where

`m=(a)/(b)\\\\y_1 = h\\\\x_1 = k`

In this case the equation of the ellipse is:

`(y^2)/(36) -(x^2)/(121)=1`

Then

`h=y_1= 0\\\\k =x_1= 0`

`m = (√(36))/(√(121))`

The asymptotes are:

`y -0=\±(√(36))/(√(121))(x-0)\\\\`

`y=\±(6)/(11)x`

see the attached image

Answer 2

D

Explanation:

The question is on asymptotes of a hyperbola

General equation for a hyperbola is; (x-h)² /a² - (y-k)²/b² =1

given y²/36 - x²/121 =1

then h=0 and k=0 thus Center of hyperbola= (0,0) at the origin.

b²=36,b⇒6 a²=121, a⇒11

Vertices of hyperbola v=( h+a, k) and (h-a, k) = (11,0) and (-11, 0)

Equation of asymptotes is given by;

y=k ± b/a (x-h) and we have found that a=11 and b=6

hence y=0 ± 6/11(x-0)

y=6/11 x and y= -6/11 x