JEE Main 2023 — Parabola Question with Solution

As $\mathrm{P}(\mathrm{b},\mathrm{c})$ lies on parabola so ${\mathrm{c}}^2=2\mathrm{ab}---(1)$

Now equation of tangent to parabola ${\mathrm{y}}^2=2\mathrm{ax}$ in point form is

${\mathrm{yy}}_1=2\mathrm{a}\frac{\left(\mathrm{x}+{\mathrm{x}}_1\right)}{2},$

Here, $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)=(\mathrm{b},\mathrm{c})$

$\Rightarrow \mathrm{yc}=\mathrm{a}(\mathrm{x}+\mathrm{b})$

For point $\mathrm{B}$, put $\mathrm{y}=0$, now $\mathrm{x}=-\mathrm{b}$

So, area of $△\mathrm{PBA},\frac{1}{2}\times \mathrm{AB}\times \mathrm{AP}=16$

$$\begin{array}{rl}& \Rightarrow \frac{1}{2}\times 2b\times c=16\\ & \\ & \Rightarrow bc=16\end{array}$$

As $\mathrm{b}$ and $\mathrm{c}$ are natural number so possible values of $(b,c)$ are $(1,16),(2,8),(4,4),(8,2)$ and $(16,1)$

Now from equation (1) $\mathrm{a}=\frac{{\mathrm{c}}^2}{2\mathrm{b}}$ and $\mathrm{a}\in \mathrm{N}$, so values of $(b,c)$ are $(1,16),(2,8)$ and $(4,4)$ now values of are 128,16 and 2 .

Hence sum of values of $a$ is 146 .