Theory of Equations 4 Question 1
1. Let $P(4,-4)$ and $Q(9,6)$ be two points on the parabola, $y^{2}=4 x$ and let $X$ be any point on the arc POQ of this parabola, where $O$ is the vertex of this parabola, such that the area of $\triangle P X Q$ is maximum. Then, this maximum area (in sq units) is
(2019 Main, 12 Jan I)
(a) $\frac{125}{2}$
(b) $\frac{75}{2}$
(c) $\frac{625}{4}$
(d) $\frac{125}{4}$
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Solution:
- Given parabola is $y^{2}=4 x$,
Since, $X$ lies on the parabola, so let the coordinates of $X$ be $\left(t^{2}, 2 t\right)$. Thus, the coordinates of the vertices of the triangle $P X Q$ are $P(4,-4), X\left(t^{2}, 2 t\right)$ and $Q(9,6)$.
$\therefore$ Area of $\triangle P X Q=\frac{1}{2}|| \begin{array}{ccc}4 & -4 & 1 \ t^{2} & 2 t & 1 \ 9 & 6 & 1\end{array}||$
$$ \begin{aligned} & =\frac{1}{2} \mid\left[4(2 t-6)+4\left(t^{2}-9\right)+1\left(6 t^{2}-18 t\right] \mid\right. \\ & =\frac{1}{2}\left|\left[8 t-24+4 t^{2}-36+6 t^{2}-18 t\right]\right| \\ & =\left|5 t^{2}-5 t-30\right|=|5(t+2)(t-3)| \end{aligned} $$
Now, as $X$ is any point on the arc $P O Q$ of the parabola, therefore ordinate of point $X, 2 t \in(-4,6) \Rightarrow t \in(-2,3)$.
$\therefore$ Area of $\triangle P X Q=-5(t+2)(t-3)=-5 t^{2}+5 t+30$
$$ [\because|x-a|=-(x-a) \text {, if } x<a] $$
The maximum area (in square units)
$$ =-\frac{25-4(-5)(30)}{4(-5)}=\frac{125}{4} $$
$[\because$ Maximum value of quadratic expression
$$ \left.a x^{2}+b x+c, \text { when } a<0 \text { is }-\frac{D}{4 a}\right] $$
