Let P(4,3) be a point on the hyperbola frac{x | vedclass

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(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.Since $P(4,3)$ lies on the hyperbola,we have $\frac{16}{a^{2}}-\frac{

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$2$

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(B) The equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.Since $P(4,3)$ lies on the hyperbola,we have $\frac{16}{a^{2}}-\frac{9}{b^{2}}=1$  $(i)$.The equation of the normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $(x_1, y_1)$ is $\frac{a^{2}x}{x_1} + \frac{b^{2}y}{y_1} = a^{2} + b^{2}$.Substituting $(x_1, y_1) = (4,3)$,the normal equation is $\frac{a^{2}x}{4} + \frac{b^{2}y}{3} = a^{2} + b^{2}$.Since the normal passes through $(16,0)$,we substitute $x=16$ and $y=0$:$\frac{a^{2}(16)}{4} + 0 = a^{2} + b^{2}$ $\Rightarrow 4a^{2} = a^{2} + b^{2}$ $\Rightarrow 3a^{2} = b^{2}$.Substitute $b^{2} = 3a^{2}$ into equation $(i)$:$\frac{16}{a^{2}} - \frac{9}{3a^{2}} = 1$ $\Rightarrow \frac{16}{a^{2}} - \frac{3}{a^{2}} = 1$ $\Rightarrow \frac{13}{a^{2}} = 1$ $\Rightarrow a^{2} = 13$.Then $b^{2} = 3(13) = 39$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^{2}}{a^{2}}} = \sqrt{1 + \frac{39}{13}} = \sqrt{1 + 3} = \sqrt{4} = 2$.

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

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