From a variable point P $(t^{2}, 2 t) (1 \leq t \leq 3)$ on the parabola $y^{2} = 4 x$ perpendicular PM is drawn to the tangent at the vertex of the parabola. If from the midpoint Q of PM, perpendicular QL is drawn to the focal chord of parabola through P (L is on the focal chord through P) then the maximum length of QL is equal to_______.

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answer is 2.7.

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Detailed Solution

$Q = (\frac{t^{2}}{2} .2 t) (y - 0) = \frac{2 t}{t^{2} - 1} (x - 1) P S : 2 t x - (t^{2} - 1) y - 2 x = 0$

Perpendicular distance from Q to PS
$P = | \frac{t^{2} - 2 t (t^{2} - 1) - 2 t}{\sqrt{4 t^{2} + {(t^{2} - 1)}^{2}}} | = \frac{t^{2}}{t^{2} + 1}$
On solving perpendicular distance $Q L = f (t) = \frac{t^{2}}{t^{2} + 1}$ so, maximum value of $f (t)$ will be at $t = 8$ which is $\frac{27}{10}$ units.

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