Let A(4, -4) and B(9, 6) be points on the par | vedclass

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(D) The equation of the parabola is $y^2 = 4x$,so $a = 1$. Let the coordinates of point $C$ be $(t^2, 2t)$.The area of $\Delta ACB$ with vertices $A(4

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$31\frac{1}{4}$

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(D) The equation of the parabola is $y^2 = 4x$,so $a = 1$. Let the coordinates of point $C$ be $(t^2, 2t)$.The area of $\Delta ACB$ with vertices $A(4, -4)$,$B(9, 6)$,and $C(t^2, 2t)$ is given by the determinant formula:$	ext{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$= \frac{1}{2} |4(6 - 2t) + 9(2t - (-4)) + t^2(-4 - 6)|$$= \frac{1}{2} |24 - 8t + 18t + 36 - 10t^2|$$= \frac{1}{2} |-10t^2 + 10t + 60| = |-5t^2 + 5t + 30|$To maximize the area,we find the derivative of $f(t) = -5t^2 + 5t + 30$ with respect to $t$ and set it to $0$:$f'(t) = -10t + 5 = 0 \implies t = \frac{1}{2}$.Substituting $t = \frac{1}{2}$ into the area expression:$	ext{Area} = |-5(\frac{1}{4}) + 5(\frac{1}{2}) + 30| = |-\frac{5}{4} + \frac{10}{4} + \frac{120}{4}| = \frac{125}{4} = 31\frac{1}{4}$ sq. units.

Let $A(4, -4)$ and $B(9, 6)$ be points on the parabola $y^2 = 4x$. Let $C$ be a point 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

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