Q51 / 155 · WBJEE 2019 · Mathematics
Let $P(4,3)$ be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=\mathrm{1}$. If the normal at $P$ intersects the $X$ -axis at (16,0) , then the eccentricity of the hyperbola is
- A$\frac{\sqrt{5}}{2}$
- Correct answerB2
- C$\sqrt{2}$
- D$\sqrt{3}$
Solution
Given hyperbola equation
$H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$P(4,3) $ lie on $H$
$\Rightarrow$ $\frac{16}{a^{2}}-\frac{9}{b^{2}}=1$
Normal equation at $P(4,3)$ for $H$
$$
a^{2} y_{1}\left(x-x_{1}\right)+b^{2} x_{1}\left(y-y_{1}\right)=0
$$
$\Rightarrow \quad 3 a^{2}(x-4)+4 b^{2}(y-3)=0$
Normal cuts the $X$ -axis at (16,0) $\Rightarrow \quad 3 a^{2}(16-4)+4 b^{2}(0-3)=0$
$\Rightarrow \quad 3 a^{2} 12+4 b^{2}(-3)=0$
$\Rightarrow$ $36 a^{2}-12 b^{2}=0$
$\Rightarrow$ $3 a^{2}-b^{2}=0$
$\Rightarrow$ $3 a^{2}=b^{2}$
$\therefore$ Eccentricity of the hyperbola is $e=\frac{\sqrt{a^{2}+b^{2}}}{a}$
$=\frac{\sqrt{a^{2}+3 a^{2}}}{a}=\frac{\sqrt{4 a^{2}}}{a}=\frac{2 a}{a}=2$
Hence, the correct answer is option: 'B'.