JEE Advanced 2018 Paper 1 Mathematics: Paragraph 2

A focal chord of $y^2=4x$ through $P(t^2,2t)$ has the other end $Q\left(\frac1{t^2},-\frac2t\right)$, since the parameters satisfy $t_1t_2=-1$. The mid-point is $(h,k)$ with $h=\frac12\left(t^2+\frac1{t^2}\right)$ and $k=t-\frac1t$. Squaring, $k^2=t^2+\frac1{t^2}-2=2h-2$. So the locus is $y^2=2(x-1)$, a parabola with vertex $(1,0)$. The other options do not satisfy $k^2=2h-2$; for instance $y^2=4(x-1)$ is what one gets by forgetting that the midpoint halves both coordinates, and $y^2=x-1$ results from dropping the factor $2$ in $k^2=t^2+t^{-2}-2$.

Tangents at $P$ and $Q$ are $ty=x+t^2$ and $-\frac yt=x+\frac1{t^2}$. Subtracting eliminates $x$: $y\left(t+\frac1t\right)=t^2-\frac1{t^2}$, so $y=t-\frac1t$ and then $x=-1$, hence $T=\left(-1,t-\frac1t\right)$ lies on the directrix. Using the determinant formula for the area with $P$, $Q$, $T$ gives $\text{Area}=\frac12\left(t+\frac1t\right)^3$. By AM-GM, $t+\frac1t\ge2$ with equality at $t=1$, so the minimum is $\frac12\cdot8=4$. Check at $t=1$: $P=(1,2)$, $Q=(1,-2)$, $T=(-1,0)$ gives area $\frac12\cdot4\cdot2=4$.