Let m 1 and m 2 be the slopes of the tangents drawn from the point P(4, 1) to the hyperbola H : y 2 25 - 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 and o

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Solution

Q.
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 $\frac{{(P Q)}^{2}}{α β}$ is equal to _________ . [2023]

Ans.
(8)

$Equation of tangent to the hyperbola \frac{y^{2}}{25} - \frac{x^{2}}{16} = 1$

$y = m x \pm \sqrt{25 - 16 m^{2}}$ ...(i)

$Equation (i) passes through (4, 1),$

$1 = 4 m \pm \sqrt{25 - 16 m^{2}}$ $\Rightarrow 1 - 4 m = \pm \sqrt{25 - 16 m^{2}}$

$\Rightarrow 1 + 16 m^{2} - 8 m = 25 - 16 m^{2} \Rightarrow 32 m^{2} - 8 m - 24 = 0$

$\Rightarrow 4 m^{2} - m - 3 = 0 \Rightarrow 4 m^{2} - 4 m + 3 m - 3 = 0$

$\Rightarrow 4 m (m - 1) + 3 (m - 1) = 0$

$\Rightarrow (4 m + 3) (m - 1) = 0 \Rightarrow m = 1, - \frac{3}{4}$

It is given that $| m_{1} |$ and $| m_{2} |$ are the slopes of tangents, so

$| m_{1} |$ $= 1$ and $| m_{2} |$ $= \frac{3}{4}$

Equations of tangents with positive slope are:

$y = x \pm \sqrt{25 - 16} \Rightarrow y = x \pm 3$

and $y = \frac{3}{4} x \pm \sqrt{25 - 16 \times \frac{9}{16}} \Rightarrow 4 y = 3 x \pm 16$

$For positive intercept on the x -axis, y = x - 3$ and $4 y = 3 x - 16$

$∴$ $α = \frac{16}{3}$ and $β = 3$

$Solve y = x - 3 and 4 y = 3 x - 16 to find their intersection point.$

$\Rightarrow x = - 4$ and $y = - 7$

$So, Q \equiv (- 4, - 7)$

${(P Q)}^{2} = {(4 + 4)}^{2} + {(1 + 7)}^{2} = 64 + 64 = 128$

$∴$ $\frac{{(P Q)}^{2}}{α β} = \frac{128}{16} = 8$

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