Solve the problem.
-The hyperbola $\frac { x ^ { 2 } } { 400 } - \frac { y ^ { 2 } } { 225 } = 1$ is shifted horizontally and vertically to obtain the hyperbola $\frac { ( x + 5 ) ^ { 2 } } { 400 } - \frac { ( y - 3 ) ^ { 2 } } { 225 } = 1$ Find the asymptotes of the new hyperbola.
A) $y - 3 = \pm \frac { 3 } { 4 } ( x + 5 )$
B) $y = \pm \frac { 3 } { 4 } x$
C) $y - 3 = \pm \frac { 4 } { 3 } ( x + 5 )$
D) $y - 3 = \pm \frac { 9 } { 16 } ( x + 5 )$

Question

Quizplus · Accepted Answer

The general equation of a hyperbola centered at point (h,k) with horizontal axis is $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$ and the equation of its asymptotes are $y-k=\pm\frac{b}{a}(x-h)$. We can observe that the given hyperbola has its center at origin(0,0) and the distance from its center to the vertices is 20 in the x direction and 15 in the y direction. Therefore, we can conclude that a=20 and b=15.

The new hyperbola is obtained by shifting 5 units to the left and 3 units up. Therefore, the new center is (-5,3) and the same equation for hyperbola that we can use is $\frac{(x+5)^2}{400}-\frac{(y-3)^2}{225}=1$. We can see that this hyperbola has the center shifted 5 units to the left of the origin and 3 units up. Therefore, we can obtain the equation of the asymptotes of the new hyperbola by substituting the values of (h,k,a,b) into the expression for asymptotes. Thus, the equation for asymptotes is $y-3=\pm\frac{3}{4}(x+5)$. Therefore, the correct choice is $\boxed{	extbf{(A)}}$.

Solve the Problem x2400−y2225=1\frac { x ^ { 2 } } { 400 } - \frac { y ^ { 2 } } { 225 } = 1400x2​−225y2​=1

Solve the Problem $\frac { x ^ { 2 } } { 400 } - \frac { y ^ { 2 } } { 225 } = 1$