Let the tangent to the parabola y 2 = 12 x at the point ( 3 ,) be perpendicular to the line 2 x + 2 y = 3 . Then the square of distance of the point (6, - 4) from the normal to the hyperbola 2 x 2 - 9 y 2 = 9 2 at its point ( 1 ,+ 2 ) is equal to

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Solution

Q.
Let the tangent to the parabola $y^{2} = 12 x$ at the point $(3, α)$ be perpendicular to the line $2 x + 2 y = 3 .$ Then the square of distance of the point (6, - 4) from the normal to the hyperbola $α^{2} x^{2} - 9 y^{2} = 9 α^{2}$ at its point $(α - 1, α + 2)$ is equal to ____ . [2023]

Ans.
(116)

$Given, line is 2 x + 2 y = 3$

$\Rightarrow y = - x + \frac{3}{2}$ ...(i)

$Slope of (i) is m = - 1$

$∴$ $Tangent at (3, α) has slope 1 .$

$Now, α^{2} = 12 (3) = 36 \Rightarrow α = 6 [∵ α = - 6 reject]$

$Equation of tangent to y^{2} = 12 x at point (3, 6) is given by$

$y - 6 = x - 3 \Rightarrow x - y + 3 = 0$

$Equation of hyperbola is α^{2} x^{2} - 9 y^{2} = 9 α^{2}$

$\Rightarrow \frac{x^{2}}{9} - \frac{y^{2}}{36} = 1$

$Equation of normal to the hyperbola at point (α - 1, α + 2),$

$i.e., (5, 8) is \frac{9 x}{5} + \frac{36 y}{8} = 45 \Rightarrow 2 x + 5 y - 50 = 0$

$Now, distance of (6, - 4) from 2 x + 5 y - 50 = 0 is given by$

$d =$ $| \frac{2 (6) - 5 (4) - 50}{\sqrt{4 + 25}} |$ $= \frac{58}{\sqrt{29}}$

$∴$ $Square of distance = {(\frac{58}{\sqrt{29}})}^{2} = 116 .$

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