Let $m_{1}$ and $m_{2}$ be the slopes of the tangents drawn from the point P(4, 1) to the hyperbola $H : \frac{y^{2}}{25} - \frac{x^{2}}{16} = 1$ . If Q is the point from which the tangents drawn to H have slopes $| m_{1} |$ and $| m_{2} |$ and they make positive intercepts $α$ and $β$ on the x-axis, then $\frac{{(PQ)}^{2}}{αβ}$ is equal to __.

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answer is 8.

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Detailed Solution

A tangent be $y = mx + \sqrt{- 16 m^{2} + 2^{5}}$ The tangent passes through (4,1)
$\Rightarrow 4 m^{2} - m - 3 = 0 \Rightarrow m = \frac{- 3}{4} or 1 | m_{1} | = \frac{3}{4}, | m_{2} | = 1$
Tangents are $y = \frac{3}{4} x - \sqrt{- 16 {(\frac{3}{4})}^{2} + 25}$ and $y = 1 . x - \sqrt{- 16 . 1^{2} + 25}$ $\Rightarrow 3 x - 4 y - 16 = 0$
$and x - y - 3 = 0$ $Solving Q = (- 4, - 7)$ $α = \frac{16}{3}, β = 3 ∴ \frac{{(PQ)}^{2}}{αβ} = \frac{128}{16} = 8$