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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Mathematics
Chapter
Parabola

JEE Advanced 2007 · Paper 1 · Mathematics · Parabola

Paragraph: Consider the circle x^2+y^2=9 and the parabola y^2=8 x ....

Paragraph: Consider the circle x2+y2=9x^2+y^2=9 and the parabola y2=8xy^2=8 x. They intersect at PP and QQ in the first and the fourth quadrants, respectively. Tangents to the circle at PP and QQ intersect the X\mathrm{X}-axis at RR and tangents to the parabola at PP and QQ intersect the X\mathrm{X}-axis at SS.Question: The radius of the circumcircle of the △PRS\triangle P R S is

Options

  1. A5
  2. B333 \sqrt{3}
  3. C323 \sqrt{2}
  4. D232 \sqrt{3}

Correct answer

B. 333 \sqrt{3}

Step-by-step solution

Equation of circumcircle of △PRS\triangle P R S is (x+1)(x-9)+y²+\lambda y=0 It will pass through (1,22)(1,2 \sqrt{2}), then −16+8+λ⋅22=0λλ=822=22\begin{aligned} -16+8+\lambda \cdot2 \sqrt{2}& =0 \\ \lambda \quad \lambda & =\frac{8}{2 \sqrt{2}}=2 \sqrt{2} \end{aligned}Equation of circumcircle is x²+y²-8 x+2 2\sqrt{2}y-9=0 Hence, its radius is 333 \sqrt{3}. ALITER Let  Let ∠PSR=θ⇒sin⁡θ=2223⇒PR=62=2R⋅sin⁡θ\begin{array}{ll}\text{ Let }& \angle P S R=\theta\\ \Rightarrow & \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}}\\ \Rightarrow & P R=6 \sqrt{2}=2 R \cdot \sin \theta \end{array} \Rightarrow \quad \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}} \Rightarrow \quadR=3R=3\sqrt{3} . \text{. }
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