Question

If the eccentricity of the standard hyperbola passing through the point $(4,6)$ is $2$, then the equation of the tangent to the hyperbola at $(4,6)$ is :

• A. $3x-2y=0$
• B. $2x-y-2=0$
• C. $2x-3y+10=0$
• D. $x-2y+8=0$

Answer 1

The correct option is B $2x-y-2=0$

Let equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$e=2\Rightarrow \sqrt{1+\frac{b^2}{a^2}}=2\Rightarrow b^2=3a^2$
The hyperbola passes through $(4,6)$
$\therefore \frac{16}{a^2}-\frac{36}{b^2}=1$
$\Rightarrow \frac{16}{a^2}-\frac{36}{3a^2}=1$
$\Rightarrow a^2=4$
So, $b^2=12$
Thus, the equation of hyperbola is $\frac{x^2}{4}-\frac{y^2}{12}=1$
$\therefore$ Equation of the tangent to the hyperbola at $(4,6)$ is $\frac{4x}{4}-\frac{6y}{12}=1$
or, $2x-y-2=0$