JEE Advanced · Mathematics · 13. Parabola
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Consider the circle \(x^2+y^2=9\) and the parabola \(y^2=8 x\). They intersect at \(P\) and \(Q\) in the first and the fourth quadrants, respectively. Tangents to the circle at \(P\) and \(Q\) intersect the \(\mathrm{X}\)-axis at \(R\) and tangents to the parabola at \(P\) and \(Q\) intersect the \(\mathrm{X}\)-axis at \(S\).Question:
The ratio of the areas of the \(\triangle P Q S\) and \(\triangle P Q R\) is
- A
\(1: \sqrt{2}\)
- B
\(1: 2\)
- C
\(1: 4\)
- D
\(1: 8\)
Answer & Solution
Correct Answer
(C)
\(1: 4\)
Step-by-step Solution
Detailed explanation
Coordinates of \(P\) and \(Q\) are \((1,2 \sqrt{2})\) and \((1,-2 \sqrt{2})\).

Area of \(\triangle P Q R=\frac{1}{2} \cdot 4 \sqrt{2} \cdot 8=16 \sqrt{2}\)
Area of \(\triangle P Q S=\frac{1}{2} \cdot 4 \sqrt{2} \cdot 2=4 \sqrt{2}\)
Ratio of area of \(\triangle P Q S\) and \(\triangle P Q R\) is \(1: 4\).

Area of \(\triangle P Q R=\frac{1}{2} \cdot 4 \sqrt{2} \cdot 8=16 \sqrt{2}\)
Area of \(\triangle P Q S=\frac{1}{2} \cdot 4 \sqrt{2} \cdot 2=4 \sqrt{2}\)
Ratio of area of \(\triangle P Q S\) and \(\triangle P Q R\) is \(1: 4\).
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