JEE Advanced 2023 · Paper 1 · Mathematics Q7

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MathematicsSingle correct+3 / -1Question 7 of 51

Let $P$ be a point on the parabola $y^2 = 4ax$ , where $a > 0$ . The normal to the parabola at $P$ meets the $x$ -axis at a point $Q$ . The area of the triangle $PFQ$ , where $F$ is the focus of the parabola, is 120. If the slope $m$ of the normal and $a$ are both positive integers, then the pair $(a, m)$ is

(2, 3)

(1, 3)

(2, 4)

(3, 4)

Answer and worked solution

Official answer: A

Take $P=(at^2,2at)$ . The normal there has slope $m=-t$ , so $t=-m$ :
$y=-tx+2at+at^3.$

Points. The normal meets the $x$ -axis at $Q=(2a+at^2,0)$ . The focus is $F=(a,0)$ .

Area.
$\text{Area}=\tfrac12\cdot FQ\cdot|2at|=\tfrac12\cdot a(1+t^2)\cdot2a|t|=a^2m(1+m^2)=120.$

Checking the options: $(2,3)$ gives $4\cdot3\cdot10=120$ . The others give $30$ , $136$ and $612$ .

Topics:Tangents and normals to conics Parabola

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JEE Advanced 2023 · Paper 1, Mathematics question 7