JEE Main 2025 — Parabola Question with Solution

To find image of $P\left(t^2,2t\right)$ $\begin{aligned} & \frac{x-t^2}{1}=\frac{y-2 t}{1}=\frac{-2\left(t^2+2 t+4\right)}{1^2+1^2}=-(t+1)^2-3 \\ & x=t^2-(t+1)^2-3=-2 t-4 \\ & y=2 t-(t+1)^2-3=-t^2-4 \\ & t=\frac{-x-4}{2} \\ & \Rightarrow y+4=-\left(\frac{-x-4}{2}\right)^2 \\ & \Rightarrow(y+4)=-\frac{(x+4)^2}{4} \\ & \Rightarrow x^2=-4 y \\ & \Rightarrow \text { Focus }(-4,-5) \end{aligned}$ $\text{ Also, }y=-5\text{ intersect }$ $\begin{array}{rl}& \therefore (-4)(-1)=(x+4{)}^2\\ & 4=(x+4{)}^2\\ & x+4=\pm 2\\ & x=-2,-6\\ & \Rightarrow d=4\\ & a=\frac{1}{2}\left|\begin{array}{ccc}1& 0& 1\\ -2& -5& 1\\ -6& -5& 1\end{array}\right|\\ & =\frac{1}{2}|[1(-5+5)+1(10-30)]|\\ & =\frac{1}{2}(20)\\ & a=10\\ & \therefore a+d=14\end{array}$