Let the length of the focal chord PQ of the parabola be 15 units. If the distance of PQ from the origin is p, then is equal to __________. [2024]

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Solution

Q.
Let the length of the focal chord PQ of the parabola $y^{2} = 12 x$ be 15 units. If the distance of PQ from the origin is p, then $10 p^{2}$ is equal to __________. [2024]

Ans.
(72)

$P Q = 15 \Rightarrow (3 {(t^{2} - \frac{1}{t^{2}}))}^{2} + (6 {(t + \frac{1}{t}))}^{2} = 225$

$\Rightarrow 9 {(t^{2} - \frac{1}{t^{2}})}^{2} + 36 {(t + \frac{1}{t})}^{2} = 225$

$\Rightarrow {(t + \frac{1}{t})}^{2} [{(t - \frac{1}{t})}^{2} + 4] = 25$

$\Rightarrow {(t + \frac{1}{t})}^{2} {(t + \frac{1}{t})}^{2} = 25 \Rightarrow {(t + \frac{1}{t})}^{4} = 25$

$\Rightarrow t + \frac{1}{t} = \pm \sqrt{5} \Rightarrow (t - \frac{1}{t}) = \pm 1$

$\Rightarrow$ Equation of PQ : $(y - 6 t) = (\frac{2 t}{t^{2} - 1}) (x - 3 t^{2})$

$\Rightarrow$ Distance from y – 6t = mx – $3 m t^{2}$ , where $m = \frac{2 t}{t^{2} - 1}$

$\Rightarrow p = \frac{| 3 m t^{2} - 6 t |}{\sqrt{1 + m^{2}}} = \frac{| (\frac{6 t}{t^{2} - 1}) |}{\sqrt{5}} = \frac{6}{\sqrt{5}}$

$\Rightarrow 10 p^{2} = \frac{10 \times 36}{5} = 72$ .

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