JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Parabola

Completing the square turns the curve into $(y-1{)}^2=2(x+1)$, a parabola with vertex $(-1,1)$ and $4a=2$. The line joining $A(1,3)$ and $B(1,-1)$ is $x=1$, and this is the chord of contact of the tangents drawn from $P$. Comparing $x=1$ with the chord of contact of a point $P(x_0,y_0)$, namely $(y-1)(y_0-1)=(x+1)+(x_0+1)$, forces $y_0=1$ and $x_0=-3$, so $P=(-3,1)$. Since $A$ and $B$ share the abscissa $x=1$, the base $AB$ is vertical with length $|3-(-1)|=4$, while the horizontal distance from $P$ to the line $x=1$ is $|1-(-3)|=4$. The triangle therefore has area $\frac{1}{2}\times 4\times 4=8$ square units.