IIT-JEE 2007 · Paper 1 · Mathematics Q61

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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 ratio of the areas of the triangles $PQS$ and $PQR$ is

1:\sqrt2

1 : 2

1 : 4

1 : 8

Answer and worked solution

Official answer: C

Intersection. $x^2+8x=9$ gives $x=1$ (the root $-9$ makes $y^2<0$ ): $P(1,2\sqrt2)$ , $Q(1,-2\sqrt2)$ .

R. The circle's tangent at P, $x+2\sqrt2\,y=9$ , meets the x-axis at $R(9,0)$ .

S. The parabola's tangent at P, $2\sqrt2\,y=4(x+1)$ , meets it at $S(-1,0)$ .

Areas. Both triangles have base $PQ=4\sqrt2$ on $x=1$ ; the heights are $1-(-1)=2$ and $9-1=8$ :
$\frac{[PQS]}{[PQR]}=\frac28=\frac14.$

(C)

Topics:Parabola Tangents and normals

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IIT-JEE 2007 · Paper 1, Mathematics question 61