IIT-JEE 2012 · Paper 1 · Mathematics Q60

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MathematicsInteger+4Question 60 of 60

Let $S$ be the focus of the parabola $y^2=8x$ and let $PQ$ be the common chord of the circle $x^2+y^2-2x-4y=0$ and the given parabola. The area of the triangle $PQS$ is

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Answer and worked solution

Official answer: 4

Intersections. Put $x=\frac{y^2}8$ in the circle:
$\frac{y^4}{64}+\frac34y^2-4y=0\implies y\left(y^3+48y-256\right)=0.$
$y=0$ or $y=4$ (the cubic is increasing, so 4 is its only real root). So $P=(0,0)$ and $Q=(2,4)$ .

Focus. $S=(2,0)$ .

Area. Base OS = 2 along the $x$ -axis, height 4 (Q is directly above S): $\frac12\cdot2\cdot4=4$ .

Answer: 4

Topics:Parabola Equation of a circle

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IIT-JEE 2012 · Paper 1, Mathematics question 60