JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Parabola Question 72 English Explanation

Given parabola $${y^2} = 4x$$

$$\therefore$$ a = 1

Here, P, S, Q points are collinear.

$$\therefore$$ Slope of PS = Slope of QS

$$ \Rightarrow {{2{t_1} - 0} \over {t_1^2 - 1}} = {{0 - 2{t_2}} \over {1 - t_2^2}}$$

$$ \Rightarrow {{2{t_1}} \over {t_1^2 - 1}} = {{2{t_2}} \over {t_2^2 - 1}}$$

$$ \Rightarrow {t_1}(t_2^2 - 1) = {t_2}(t_1^2 - 1)$$

$$ \Rightarrow t_2^2{t_1} - {t_1} = t_1^2{t_2} - {t_2}$$

$$ \Rightarrow t_2^2{t_1} - t_1^2{t_2} - {t_1} + {t_2} = 0$$

$$ \Rightarrow {t_1}{t_2}({t_2} - {t_1}) + ({t_2} - {t_1}) = 0$$

$$ \Rightarrow ({t_2} - {t_1})({t_1}{t_2} + 1) = 0$$

As $${t_2} - {t_1} \ne 0$$

$$\therefore$$ $${t_1}{t_2} + 1 = 0$$

$${t_1}{t_2} = - 1$$

Now, lenght of PQ

$$ = \sqrt {{{\left( {t_1^2 - t_2^2} \right)}^2} + {{\left( {2{t_1} - 2{t_2}} \right)}^2}} $$

$$ = \sqrt {{{\left( {{t_1} + {t_2}} \right)}^2}{{\left( {{t_1} - {t_2}} \right)}^2} + 4{{\left( {{t_1} - {t_2}} \right)}^2}} $$

$$ = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} + {t_2}} \right)}^2} + 4} $$

$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2{t_1}{t_2} + 4} $$

$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2( - 1) + 4} $$

$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 + 2} $$

$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2( - 1)} $$

$$ = \left( {{t_1} - {t_2}} \right)\sqrt {t_1^2 + t_2^2 - 2{t_1}{t_2}} $$

$$ = \left( {{t_1} - {t_2}} \right)\sqrt {{{\left( {{t_1} - {t_2}} \right)}^2}} $$

$$ = {\left( {{t_1} - {t_2}} \right)^2}$$

Given, length of $$PQ = {\left( {{t_1} - {t_2}} \right)^2} = 6.25$$

$$ \Rightarrow {t_1} - {t_2} = 2.5$$

Now, Area of $$\Delta OPQ$$

$$ = \left| {{1 \over 2}\left| {\matrix{ {t_1^2} & {2{t_1}} & 1 \cr {t_2^2} & {2{t_2}} & 1 \cr 0 & 0 & 1 \cr } } \right|} \right|$$

$$ = \left| {{1 \over 2}\left( {2{t_2}t_1^2 - 2{t_1}t_2^2} \right)} \right|$$

$$ = \left| {{1 \over 2} \times 2{t_1}{t_2}\left( {{t_1} - {t_2}} \right)} \right|$$

$$ = \left| {{t_1}{t_2} \times \left( {{t_1} - {t_2}} \right)} \right|$$

$$ = \left| { - 1 \times 2.5} \right|$$

$$ = 2.5$$

$$\therefore$$ 10$$\Delta$$OPQ =

$$ = 10 \times {{25} \over {10}} = 25$$

Parabola

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