JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Application of Derivatives

Differentiating $x^2+2x-4y+9=0$ gives $2x+2-4y^{\prime }=0$, so $y^{\prime }=\frac{x+1}{2}=1$ at $P(1,3)$. The tangent $y-3=(x-1)$, i.e. $y=x+2$, meets the $y$-axis at $A(0,2)$. The line through $P$ parallel to $x-3y=6$ has slope $\frac{1}{3}$: $y-3=\frac{1}{3}(x-1)$, i.e. $x=3y-8$. Substituting into $y^2=4x$ gives $y^2-12y+32=0$, so $y=4$ or $y=8$, giving the points $(4,4)$ and $(16,8)$. Only $(16,8)$ satisfies $2x-3y=8$, so $B=(16,8)$. Hence $(AB{)}^2=(16-0{)}^2+(8-2{)}^2=256+36=292$.