JEE Advanced · Mathematics · 13. Parabola
Let \(S\) be the focus of the parabola \(y^{2}=8 \mathrm{x}\) and let \(P Q\) be the common chord of the circle \(x^{2}+y^{2}-2 x-4 y=0\) and the given parabola. The area of the triangle \(P Q S\) is
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(B) 4
Step-by-step Solution
Detailed explanation
Given parabola \(y^{2}=8 x\)
and circle \(x^{2}+y^{2}-2 x-4 y=0\) pass through the origin
\(\therefore\) One end of common chord PQ is origin. Say \(\mathrm{P}(0,0)\)
Let \(\mathrm{Q}\) be the point \(\left(2 t^{2}, 4 t\right)\), then it will satisfy the equation of circle.
\(\therefore \quad 4 t^{4}+16 t^{2}-4 t^{2}-16 t=0\)
\(\Rightarrow t^{4}+3 t^{2}-4 t=0 \Rightarrow t\left(t^{3}+3 t-4\right)=0\)
\(\Rightarrow t(t-1)\left(t^{2}+t-4\right)=0 \Rightarrow t=0\) or 1
For \(t=0\), we get point \(P\), therefore \(t=1\) gives point \(Q\) as \((2,4)\).
We also observe here that \(\mathrm{P}(0,0)\) and \(\mathrm{Q}(2,4)\) are end points of diameter of the given circle and focus of the parabola is the point \(S(2,0)\).
\(\therefore \quad\) area \((\Delta \mathrm{PQS})=\frac{1}{2} \times P S \times Q S=\frac{1}{2} \times 2 \times 4=4\)
and circle \(x^{2}+y^{2}-2 x-4 y=0\) pass through the origin
\(\therefore\) One end of common chord PQ is origin. Say \(\mathrm{P}(0,0)\)
Let \(\mathrm{Q}\) be the point \(\left(2 t^{2}, 4 t\right)\), then it will satisfy the equation of circle.
\(\therefore \quad 4 t^{4}+16 t^{2}-4 t^{2}-16 t=0\)
\(\Rightarrow t^{4}+3 t^{2}-4 t=0 \Rightarrow t\left(t^{3}+3 t-4\right)=0\)
\(\Rightarrow t(t-1)\left(t^{2}+t-4\right)=0 \Rightarrow t=0\) or 1
For \(t=0\), we get point \(P\), therefore \(t=1\) gives point \(Q\) as \((2,4)\).
We also observe here that \(\mathrm{P}(0,0)\) and \(\mathrm{Q}(2,4)\) are end points of diameter of the given circle and focus of the parabola is the point \(S(2,0)\).
\(\therefore \quad\) area \((\Delta \mathrm{PQS})=\frac{1}{2} \times P S \times Q S=\frac{1}{2} \times 2 \times 4=4\)
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