Question bank/JEE Main/Mathematics

Let A(4, -4) and B(9, 6) be points on the parabola, y^2 = 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

A$30\frac{1}{2}$B$31\frac{3}{4}$C$31\frac{1}{4}$D32

Show answer and solution▾

$\Delta$ = 30 + 5t - 5t^2
$\frac{d\Delta}{dt} = 0 \Rightarrow 5 - 10t = 0 \Rightarrow t = \frac{1}{2}$
$\frac{d^{2}\Delta}{dt^{2}} = - 10 < 0\therefore$
$C\left( \frac{1}{4},1 \right)\ so\ \Delta = 30 + \frac{5}{2} - \frac{5}{4} = 31\frac{1}{4}$

Source: Grass-G / M-22 Mathematics Paper 19 Dec Eng.docx