Let P(6, 3) be a point on the hyperbola frac{ | vedclass

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(B) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2x}{x_1} + \frac{b^

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$\sqrt{\frac{3}{2}}$

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(B) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is given by $\frac{a^2x}{x_1} + \frac{b^2y}{y_1} = a^2 + b^2$.Substituting $(x_1, y_1) = (6, 3)$,we get $\frac{a^2x}{6} + \frac{b^2y}{3} = a^2 + b^2$.Since the normal passes through $(9, 0)$,we substitute $x = 9$ and $y = 0$:$\frac{a^2(9)}{6} + 0 = a^2 + b^2$$\frac{3a^2}{2} = a^2 + b^2$$\frac{1}{2}a^2 = b^2 \Rightarrow a^2 = 2b^2$.The eccentricity $e$ of the hyperbola is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.Substituting $a^2 = 2b^2$:$e = \sqrt{1 + \frac{b^2}{2b^2}} = \sqrt{1 + \frac{1}{2}} = \sqrt{\frac{3}{2}}$.

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 point $P$ intersects the $x$-axis at $(9, 0)$,then the eccentricity of the hyperbola is:

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