Parabola Ques 18
- Let $S$ be the focus of the parabola $y^{2}=8 x$ and $P Q$ be the common chord of the circle $x^{2}+y^{2}-2 x-4 y=0$ and the given parabola. The area of $\triangle P Q S$ is
(2012)
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Solution:
- PLAN Parametric coordinates for $y^{2}=4 a x$ are $\left(a t^{2}, 2 a t\right)$.
Description of Situation As the circle intersects the parabola at $P$ and $Q$. Thus, points $P$ and $Q$ should satisfy the circle.
$$ \begin{array}{rlrl} & & P\left(2 t^{2}, 4 t\right) \text { should lie on } x^{2}+y^{2}-2 x-4 y=0 \\ \Rightarrow & 4 t^{4}+16 t^{2}-4 t^{2}-16 t & =0 \\ \Rightarrow & 4 t^{4}+12 t^{2}-16 t & =0 \\ \Rightarrow & & 4 t\left(t^{3}+3 t-4\right) & =0 \\ \Rightarrow & & 4 t(t-1)\left(t^{2}+t+4\right) & =0 \\ & \therefore & t & =0.1 \end{array} $$
$\Rightarrow P(2,4)$ and $P Q$ is the diameter of the circle.
Thus, area of $\triangle P Q S=\frac{1}{2} \cdot P S \times P Q=\frac{1}{2} \cdot(2) \cdot(4)=4$
BETA
