Question

Which equations represent the asymptotes of the hyperbola (x-1)^2/36-(y-2)^2/64=1 ?

Answer 1

Answer:

$4x+3y=10$ and  $4x-3y=-2$

Step-by-step explanation:

The asymptote equation of a translate hyperbola with equation $\frac{(x-h)^2}{b^2}-\frac{(y-k)^2}{a^2}=1$ is $y=\pm\frac{b}{a}(x-h)+k$

The given hyperbola has equation: $\frac{(x-1)^2}{36}-\frac{(y-2)^2}{64}=1$

Or $\frac{(x-1)^2}{6^2}-\frac{(y-2)^2}{8^2}=1$

Comparing to $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

We have  a=6 and b=8, h=1 and k=2.

We substitute this values to get:

$y=\pm\frac{6}{8}(x-1)+2$

$y=\pm\frac{4}{3}(x-1)+2$

$3y=\pm4(x-1)+6$

We split the plus or minus sign to get:

$3y=4x-4+6$

$3y=4x+2$

$3y-4x=2$

Or

$3y=-4(x-1)+6$

$3y=-4x+4+6$

$3y+4x=10$