Question

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 $△ACB$ is maximum. Then, the area (in sq. units) of $△ACB,$ is:

• A. $31\frac{1}{4}$
• B. $30\frac{1}{2}$
• C. $32$
• D. $31\frac{3}{4}$

Answer 1

The correct option is A $31\frac{1}{4}$

Consider the point $C(t^2,2t)$ on parabola $y^2=4x$


$\therefore$ Area of $△ABC=\frac{1}{2}\left|\left|\begin{array}{ccc}{t}^2& 2t& 1\\ 9& 6& 1\\ 4& -4& 1\end{array}\right|\right|$
$=\frac{1}{2}\left|t^2\right(6+4)-2t(9-4)+1(-36-24\left)\right|$
$=5|t^2-t-6|$
which is maximum at $t=\frac{1}{2}$
Maximum area $=5|{\left(\frac{1}{2}\right)}^2-\frac{1}{2}-6|=\frac{125}{4}=31\frac{1}{4}$ sq. units