Question

The eccentricity of a hyperbola whose transverse axis is along $x-$ axis and passes through the point $(6,4)$ is $\sqrt{\frac{5}{3}}$. The equation of tangent to this hyperbola at $(6,4)$ is

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

Answer 1

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

$e=\sqrt{\frac{5}{3}}\Rightarrow e^2=1+\frac{b^2}{a^2}$
$\Rightarrow 2a^2=3b^2\cdots \left(i\right)$
Assuming the equation of the hyperbola be,
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Putting the point $(6,4)$ in the equation of hyperbola,
$\frac{36}{a^2}-\frac{16}{b^2}=1$
Using equation (i),
$\Rightarrow a^2=12\Rightarrow b^2=8$

$\frac{x^2}{12}-\frac{y^2}{8}=1\Rightarrow \frac{x}{6}-\frac{y{y}^{\prime }}{4}=0$
Putting the point $(6,4)$
$\frac{dy}{dx}=1$
Equation of tangent is,
$(y-4)=1(x-6)$
$\Rightarrow x-y-2=0$