Question

What will be the equation of the asymptote of the hyperbola
$\frac{(x-7{)}^2}{49}-\frac{(y-8{)}^2}{36}=1$
slope of asymptote $=\frac{6}{7}$ with respect to the transverse axis?

• A. $6x-7y-14=0$
• B. $6x+7y+14=0$
• C. $6x-7y+14=0$
• D. $6x+7y-14=0$

Answer 1

The correct option is C $6x-7y+14=0$

Given: $\text{Hyperbola}\frac{(x-7{)}^2}{49}-\frac{(y-8{)}^2}{36}=1$;
slope of asymptote $=\frac{6}{7}$ with respect to the transverse axis

To find: Equation of asymptote.

We know that, the asymptotes of a hyperbola always pass through the centre of the hyperbola.

The standard equation of a translated hyperbola is, $\frac{(x-h{)}^2}{a^2}-\frac{(y-k{)}^2}{b^2}=1$ centered at $(h,k)$.

On comparing the above equation with the given hyperbola equation, we have, $h=7,k=8$.

Hence, the centre is at $(7,8)$.

Now, the equation of asymptote can be found using the equation of straight line $(y-y_1)=m(x-x_1)$.
$\Rightarrow m=\frac{6}{7}$

Substituting all the values in the straight line equation,
$\Rightarrow (y-8)=\frac{6}{7}(x-7)$

$\Rightarrow 7(y-8)=6(x-7)$

$\Rightarrow 7y-56=6x-42$

$\Rightarrow 6x-7y+14=0.$