Question

Let $P(4,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at $P$ intersects the X-axis at $(16,0)$, then the eccentricity of the hyperbola is

• A. $\frac{\sqrt{5}}{2}$
• B. $2$
• C. $\sqrt{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is B $2$

Slope of tangent at $(4,3)$
$\frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}=\frac{4b^2}{3a^2}$
Slope of normal at $(4,3)$ is $\frac{-3a^2}{4b^2}$

Equation of normal at $(4,3)$ is$y=\frac{-3a^2}{4b^2}x+c$
$(16,0)$ and $(4,3)$ lies on this line
$\Rightarrow 0=\frac{-3a^2}{4b^2}\cdot 16+c$
$\Rightarrow c=\frac{12a^2}{b^2}$
$\Rightarrow y=\frac{-3a^2}{4b^2}x+\frac{12a^2}{b^2}$

At $(4,3)$
$3=\frac{-3a^2}{4b^2}\cdot 4+\frac{12a^2}{b^2}$
$\Rightarrow \frac{b^2}{a^2}=3$
$\Rightarrow e^2=1+\frac{b^2}{a^2}$
$\Rightarrow e=2$