JEE MAIN - Mathematics (2019 - 9th January Evening Slot - No. 6)
Let A(4, $$-$$ 4) and B(9, 6) be points on the parabola, y2 = 4x. Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of $$\Delta $$ACB is maximum. Then, the area (in sq. units) of $$\Delta $$ACB, is :
$$31{1 \over 4}$$
$$30{1 \over 2}$$
32
$$31{3 \over 4}$$
Explanation
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$$\Delta ABC = {1 \over 2}\left| {\matrix{ 4 & { - 4} & 1 \cr 9 & 6 & 1 \cr {{t^2}} & {2t} & 1 \cr } } \right|$$
D = 60 + 10t $$-$$ 10t2
$${{d\Delta } \over {dt}} = 0 \Rightarrow t = {1 \over 2}$$
$${{{d^2}\Delta } \over {d{t^2}}} = - 20 < 0$$
$$ \therefore $$ max at $$t = {1 \over 2}$$
max area $$\Delta = 65 - {5 \over 2}$$
$$ = {{125} \over 2} = 31{1 \over 4}$$
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