Let , be the lines passing through the point P(1, 0) and touching the parabola . Let Q and R be the points on the lines and such that the is an isosceles triangle with base QR. If the slopes of the lines QR are and , then is equal to __________.

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Solution

Q.
Let $L_{1}$ , $L_{2}$ be the lines passing through the point P(0, 1) and touching the parabola $9 x^{2} + 12 x + 18 y - 14 = 0$ . Let Q and R be the points on the lines $L_{1}$ and $L_{2}$ such that the $△ P Q R$ is an isosceles triangle with base QR. If the slopes of the lines QR are $m_{1}$ and $m_{2}$ , then $16 (m_{1}^{2} + m_{2}^{2})$ is equal to __________. [2024]

Ans.
(68)

We have $9 x^{2} + 12 x + 18 y - 14 = 0$

$\Rightarrow 9 x^{2} + 12 x + 4 + 18 y - 18 = 0$

$\Rightarrow {(3 x + 2)}^{2} = - 18 y + 18$

$\Rightarrow {(x + \frac{2}{3})}^{2} = - 2 (y - 1)$

Vertex of parabola is $(\frac{- 2}{3}, 1)$

Now, equation of line passing through (0, 1) is given by y = mx + 1.

Since the line is touching the parabola, so we have

${(x + \frac{2}{3})}^{2} = - 2 m x$

$\Rightarrow {(3 x + 2)}^{2} = - 18 m x$

$\Rightarrow 9 x^{2} + (12 + 18 m) x + 4 = 0$

Discriminant of this quadratic equation must be zero.

$\Rightarrow 4 {(6 + 9 m)}^{2} = 4 (36)$

$\Rightarrow 6 + 9 m = 6 or - 6$

$\Rightarrow m = 0, \frac{- 4}{3}$

$∴ \tan θ = \frac{- 4}{3}$

$\Rightarrow \frac{2 \tan θ / 2}{1 - \tan^{2} θ / 2} = \frac{- 4}{3}$

$\Rightarrow - 4 + 4 \tan^{2} θ / 2 - 6 \tan θ / 2 = 0$

$\Rightarrow 2 \tan^{2} θ / 2 - 3 \tan θ / 2 - 2 = 0$

$\Rightarrow (\tan \frac{θ}{2} - 2) (2 \tan \frac{θ}{2} + 1) = 0$

$\Rightarrow \tan \frac{θ}{2} = 2, \frac{- 1}{2}$ ... (i)

Now, in $△ P Q R, ∠ P = θ s o ∠ Q = ∠ R = 90 - \frac{θ}{2}$

$∴ m_{Q R} = \tan (90 + \frac{θ}{2}) = - c o t \frac{θ}{2} = \frac{- 1}{2}, 2$ [Using (i)]

$\Rightarrow m_{1} = \frac{- 1}{2}, m_{2} = 2$

$∴ 16 (m_{1}^{2} + m_{2}^{2}) = 16 (\frac{1}{4} + 4) = 68$ .

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