JEE Main 2023 — Parabola Question with Solution

Given,

Equation of the curves $x=2y^2$ and $x=1+y^2$ 

Now for parabola $y^2=\frac{x}{2}$, equation of tangent is given by $y=mx+\frac{1}{8m}$

Which is also tangent to parabola  $y^2+1=x$,

So, $y=mx+\frac{1}{8m}$ will satisfy the curve $y^2+1=x$,

Hence, ${\left(\mathrm{mx}+\frac{1}{8m}\right)}^2+1=x$

$\Rightarrow \left(m^2{x}^2+\frac{1}{64m^2}+\frac{x}{4}\right)+1=x$

$\Rightarrow m^2{x}^2-\frac{3x}{4}+1+\frac{1}{64m^2}=0$

Now making discriminant zero we get,

$\mathrm{D}=0\Rightarrow {\left(\frac{3}{4}\right)}^2-4\times m^2\times \left(1+\frac{1}{64{\mathrm{m}}^2}\right)$

$\Rightarrow \left(\frac{9}{16}\right)-\left(4m^2+\frac{1}{16}\right)=0$

$\Rightarrow m=\frac{1}{2\sqrt{2}}$

Hence, the equation of tangent is $x-2\sqrt{2}y+1=0$

Now distance of point $\left(6,-2\sqrt{2}\right)$ from the tangent will be, $d=\left|\frac{6+8+1}{\sqrt{9}}\right|=5$