Answer: Option D
[tex]y=\frac{6}{11}x[/tex]
and
[tex]y=-\frac{6}{11}x[/tex]
Step-by-step explanation:
We know that the general equation of a hyperbola with transverse vertical axis has the form:
[tex]\frac{(y-h)^2}{a^2}-\frac{(x-k)^2}{b^2} =1[/tex]
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
[tex]y -y_1=\±m(x-x_1)\\\\[/tex]
Where
[tex]m=\frac{a}{b}\\\\y_1 = h\\\\x_1 = k[/tex]
In this case the equation of the ellipse is:
[tex]\frac{y^2}{36} -\frac{x^2}{121}=1[/tex]
Then
[tex]h=y_1= 0\\\\k =x_1= 0[/tex]
[tex]m = \frac{\sqrt{36}}{\sqrt{121}}[/tex]
The asymptotes are:
[tex]y -0=\±\frac{\sqrt{36}}{\sqrt{121}}(x-0)\\\\[/tex]
[tex]y=\±\frac{6}{11}x[/tex]
see the attached image
