JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Parabola

The parabola $y^2=8x$ has points $\left(\frac{y^2}{8},y\right)$, and the circle $x^2+y^2+12y+35=0$ rewrites as $x^2+(y+6{)}^2=1$, a circle of centre $C(0,-6)$ and radius $1$. The shortest distance between the curves is measured along the line joining $C$ to the nearest point of the parabola, so minimise the distance from $C$ to a point $P\left(\frac{y^2}{8},y\right)$. With $D^2=\frac{y^4}{64}+(y+6{)}^2$, setting $\frac{d(D^2)}{dy}=\frac{y^3}{16}+2(y+6)=0$ gives $y^3+32y+192=0$, which has the root $y=-4$. The nearest point is $P(2,-4)$, and $CP=\sqrt{2^2+(-4+6{)}^2}=\sqrt{8}=2\sqrt{2}$. Subtracting the radius, the shortest distance between the curves is $2\sqrt{2}-1$.