JEE Advance - Mathematics (2023 - Paper 1 Online - No. 7)
Let $P$ be a point on the parabola $y^2=4 a x$, where $a>0$. The normal to the parabola at $P$ meets the $x$-axis at a point $Q$. The area of the triangle $P F Q$, 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
$(2,3)$
$(1,3)$
$(2,4)$
$(3,4)$
Explanation
$$ y^2=4 a x $$
Equation of normal
$$ y=m x-2 a m-a m^3 $$
Point of contact
$$ P\left(a m^2,-2 a m\right) $$
and Point $Q\left(2 a+a m^2, 0\right)$
Area of
$\begin{aligned} \triangle P F Q & =\frac{1}{2} \times\left|a+a m^2\right||-2 a m| \\\\ 120 & =a^2\left(1+m^2\right) m \\\\ a & =2, m=3\end{aligned}$
Satisfies the equation $(1)$, hence $(2,3)$ will be the correct answer.
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