Question

Consider the parabola $y^2=4x$;$A(4,-4);B(9,6)$ be two fixed points on the parabola. Let $C$ be a moving point on the parabola between $A$ and $B$ such that the area of the triangle $ABC$ is maximum. The coordinate of $C$ is

• A. $(\frac{1}{4},1)$
• B. $(4,4)$
• C. $3,2\sqrt{3})$
• D. $3,-2\sqrt{3})$

Answer 1

The correct option is A $(\frac{1}{4},1)$

Parabola: $y^2=4x.........\left(i\right)(a=1)$

Point of Parabola $A(4,-4)$ and $B:(9,6)$

Let $C$ be $({at}^2,2at):C(t^2,2t)$

Now, area of triangle $ABC=\frac{1}{2}\left|\begin{array}{cc}4& -4\\ 9& 6\\ t^2& 2t\\ 4& -4\end{array}\right|$

Area of $△ABC=\frac{1}{2}(24+36+18t-6t^2-{4t}^2-8t)$

$=\frac{1}{2}(60-{10t}^2+10t)$

$=30-5t^2+5t.......\left(ii\right)$

For maximum value of $\left(ii\right)$ $=>-5\left(2t\right)+5=0$ $(\frac{d{\left(2\right)}^2}{dt}=0)$

$=>t=\frac{1}{2}$

So, $C$ becomes, $\left(\right(\frac{1}{2}{)}^2,2\left(\frac{1}{2}\right))$

$C$:$(\frac{1}{4},1)$