Let $P(6, 3)$ be a point on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$. If the normal at the point $P$ intersects the $x$-axis at $(9, 0)$, then the eccentricity of the hyperbola is

Let P(6, 3) be a point on the hyperbola x²/a² - y²/b² = 1. If the normal at the point P intersects the x-axis at (9, 0), then the eccentricity of the hyperbola is

$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1$

$P(6,3) \text{ lies on it}$

$\text{Normal at } P \text{ meets } x\text{-axis at } (9,0)$

$\text{Find eccentricity } e = \sqrt{1+\dfrac{b^{2}}{a^{2}}}$

$\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1,\qquad P(6,3)$

$\text{Normal at } (x_1,y_1):\quad \dfrac{a^{2}x}{x_1}+\dfrac{b^{2}y}{y_1}=a^{2}+b^{2}$

$y=0:\ \dfrac{a^{2}x}{x_1}=a^{2}+b^{2} \Rightarrow x=\dfrac{x_1(a^{2}+b^{2})}{a^{2}}$

$9=\dfrac{6(a^{2}+b^{2})}{a^{2}} \Rightarrow \dfrac{a^{2}+b^{2}}{a^{2}}=\dfrac{3}{2}$

$\Rightarrow \dfrac{b^{2}}{a^{2}}=\dfrac{1}{2}$

$e^{2}=1+\dfrac{b^{2}}{a^{2}}=1+\dfrac{1}{2}=\dfrac{3}{2}$

$e=\sqrt{\dfrac{3}{2}}\ \Rightarrow\ \text{option (2)}$

$\dfrac{36}{a^{2}}-\dfrac{9}{b^{2}}=1,\quad b^{2}=\dfrac{a^{2}}{2} \Rightarrow a^{2}=18,\ b^{2}=9$

$\boxed{\text{Answer: } \sqrt{\dfrac{3}{2}} \text{ — option (2)}}$