JEE Main 2023 — Parabola Question with Solution

Given,

$S$ be the set of all $a\in N$ such that the area of the triangle formed by the tangent at the point $P\left(b,c\right), b, c\in N$,  on the parabola $y^2=2ax$

And the lines $x=b,y=0$ is $16{\mathrm{unit}}^2$

Now plotting the diagram we get,

Tangent at $P\left(b,c\right)$ will be $ax-cy+ab=0$ and it meets $x$-axis at $\left(-b,0\right)$

Now area of triangle $\left(PAB\right)=\frac{1}{2}\left(2b\right)\left(c\right)=16\Rightarrow bc=16$

So, $\left(b,c\right)=\left(1,16\right),\left(2,8\right),\left(4,4\right),\left(8,2\right),\left(16,1\right)$ as both belongs to natural number,

Now $\left(b,c\right)$ lie on parabola hence satisfy it

$c^2=2ab\Rightarrow c^2=2a\left(\frac{16}{c}\right)\Rightarrow a=\frac{c^3}{32}$

Now when possible value of $c$ is $c=4,8,16\Rightarrow a=2,16,128$

Hence, $\sum _{a\in S}a=2+16+128=146$.