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


Let $A(4,-4)$ and $B(9,6)$ be points on the parabola, $\mathrm{y}^{2}+4 \mathrm{x}$. Let $\mathrm{C}$ be chosen on the arc $\mathrm{AOB}$ of the parabola, where $O$ is the origin, such that the area of $\triangle A C B$ is maximum. Then, the area (in sq. units) of $\triangle \mathrm{ACB}$, is:

  1. $31 \frac{3}{4}$

  2. 32

  3. $30 \frac{1}{2}$

  4. $31 \frac{1}{4}$

Correct Option: , 4


Area $=5\left|\mathrm{t}^{2}-\mathrm{t}-6\right|=5\left|\left(\mathrm{t}-\frac{1}{2}\right)^{2}-\frac{25}{4}\right|$

is maximum if $\mathrm{t}=\frac{1}{2}$

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