IIT-JEE 2007 · Paper 1 · Mathematics Q62

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MathematicsSingle correct+4 / -1Question 62 of 66

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Paragraph for Question Nos. 61 to 63

Consider the circle $x^2+y^2=9$ and the parabola $y^2=8x$ . They intersect at $P$ and $Q$ in the first and the fourth quadrants, respectively. Tangents to the circle at $P$ and $Q$ intersect the $x$ -axis at $R$ and tangents to the parabola at $P$ and $Q$ intersect the $x$ -axis at $S$ .

The radius of the circumcircle of the triangle $PRS$ is

3\sqrt3

3\sqrt2

2\sqrt3

Answer and worked solution

Official answer: B

Sides of PRS with $P(1,2\sqrt2)$ , $R(9,0)$ , $S(-1,0)$ : $SR=10$ , $PS=\sqrt{4+8}=2\sqrt3$ , $PR=\sqrt{64+8}=6\sqrt2$ .

Area. $\frac12\cdot10\cdot2\sqrt2=10\sqrt2$ .

Circumradius.
$R=\frac{abc}{4\Delta}=\frac{10\cdot2\sqrt3\cdot6\sqrt2}{40\sqrt2}=3\sqrt3.$

(Check: $\vec{PS}\cdot\vec{PR}=(-2)(8)+8=-8\ne0$ , so the triangle is not right-angled and the formula is needed.)

(B)

Topics:Parabola Tangents and normals

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IIT-JEE 2007 · Paper 1, Mathematics question 62