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Let $\mathrm{m}_1$, and $\mathrm{m}_2$ be the slopes of the tangents drawn from the point $\mathrm{P}(4,1)$ to the hyperbola $\mathrm{H}: \frac{y^2}{25}-\frac{x^2}{16}=1$ . If $\mathrm{Q}$ is the point from which the tangents drawn to $\mathrm{H}$ have slopes $\left|\mathrm{m}_1\right|$ and $\left|\mathrm{m}_2\right|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$ -axis, then $\mathrm{ \frac{(P Q)^2}{\alpha \beta}}$ is equal to

Option: 1
8

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

Equation of tangent to the hyperbola $\mathrm{\frac{y^2}{a^2}-\frac{x^2}{b^2}=1}$
$\mathrm{y=m x \pm \sqrt{a^2-b^2 m^2}}$
passing through $(4,1)$
$\begin{aligned} & 1=4 \mathrm{~m} \pm \sqrt{25-16 \mathrm{~m}^2} \Rightarrow 4 \mathrm{~m}^2-\mathrm{m}-3=0 \\ & \Rightarrow \mathrm{m}=1, \frac{-3}{4} \end{aligned}$

Equation of tangent with positive slopes $1 \& \frac{3}{4}$
$\left.\begin{array}{c} 4 y=3 x-16 \\ y=x-3 \end{array}\right\} \text { with positive intercept on } x \text {-axis. }$
$\alpha=\frac{16}{3}, \beta=3$
Intersection points:

$\mathrm{Q:\left ( -4,-7 \right )}$

$\mathrm{P:\left ( 4,1 \right )}$

$\mathrm{PQ^{2}:\left ( 128 \right )}$

$\mathrm{\frac{PQ^{2}}{\alpha \beta }=\frac{128}{16}=8}$

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sudhir.kumar

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