JEE Main 2025 — Application Of Derivatives Question with Solution

Equation of the Normal to the Parabola:

The equation for the normal to the parabola $y^2=8x$ can be expressed as:

$y=mx-2am-a{m}^3\text{where }a=2$

Simplifying, we have:

$y=mx-4m-2m^3$

Center and Radius of the Circle:

The given second equation $x^2+y^2+12y+35=0$ can be rewritten to find the center and radius of the circle:

Rewrite it as $x^2+(y+6{)}^2=1$.

Thus, the center of the circle is $C(0,-6)$ and the radius $r=1$.

Finding the Slope of the Normal:

To find the point $P$ on the parabola where the normal meets, equate:

$-6=-4m-2m^3$

Solving for $m$ gives:

$m=1$

Determine Point $P$ on the Parabola:

Substituting $m=1$ back to find $P\left(a{m}^2,-2am\right)$:

$P(2,-4)$

Calculate the Shortest Distance:

The shortest distance from point $P(2,-4)$ to the center $C(0,-6)$ minus the radius is:

$CP-r=\sqrt{(2-0{)}^2+(-4+6{)}^2}-1=\sqrt{4+4}-1$

$=\sqrt{8}-1=2\sqrt{2}-1$

Thus, the shortest distance between the given curves is $2\sqrt{2}-1$.