P(6, 3) is a point on the hyperbola frac{x^2} | vedclass

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(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Differentiating with respect to $x$,we get $\frac{2x}{a^2} - \frac{2y}{b^

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$\sqrt{\frac{5}{3}}$

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(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Differentiating with respect to $x$,we get $\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = \frac{b^2 x}{a^2 y}$.At $P(6, 3)$,the slope of the tangent is $m_t = \frac{b^2(6)}{a^2(3)} = \frac{2b^2}{a^2}$.The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{a^2}{2b^2}$.The equation of the normal at $P(6, 3)$ is $y - 3 = -\frac{a^2}{2b^2}(x - 6)$.Since the normal passes through $(10, 0)$,we have $0 - 3 = -\frac{a^2}{2b^2}(10 - 6)$.$-3 = -\frac{a^2}{2b^2}(4)$ $\Rightarrow 3 = \frac{2a^2}{b^2}$ $\Rightarrow \frac{a^2}{b^2} = \frac{3}{2}$.The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{2}{3} = \frac{5}{3}$.Therefore,$e = \sqrt{\frac{5}{3}}$.

$P(6, 3)$ is 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 $(10, 0)$,then the eccentricity of the hyperbola is

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