JEE Main 2023 — Application of Derivatives Question with Solution

Given,

The tangent to the curve $x^2+2x-4y+9=0$ at the point $P\left(1,3\right)$ on it meet the $y-$axis at $A$,

So, tangent to circle at point $P\left(1,3\right)$ is given by,

 $x+\left(x+1\right)-2\left(y+3\right)+9=0$

$\Rightarrow 2x-2y=-4$

So, on $y-$axis, $A\left(0,2\right)$

And  the line passing through $P$ and parallel to the line $x-3y=6$ meet the parabola $y^2=4x$ at $B$,

So, the equation of line through $P$ will be $y-3=\frac{1}{3}\left(x-1\right)$

$\Rightarrow 3y=x+8$

Now finding intersection of $y^2=4x \& 3y=x+8$ we get,

$B=\left(4,4\right) \& \left(16,8\right)$, 

Also $B$ lies on the line $2x-3y=8,$ so $\left(4,4\right)$ will not satisfy,

Hence, $B=\left(16,8\right)$ 

So, ${\left(AB\right)}^2={\left(0-16\right)}^2+{\left(2-8\right)}^2=256+36=292$