JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Parabola

The parabola is $y^2=2x-3$, which can be written as $y^2=2(x-\frac{3}{2})$. The chord of contact from $R(0,1)$ is $y\cdot 1=(x+0)-3$, giving $y=x-3$. Substituting into the parabola: $(x-3{)}^2=2x-3$, so $x^2-8x+12=0$, giving $x=2$ or $x=6$. Thus $P(2,-1)$ and $Q(6,3)$. The slope of PQ is $\frac{3-(-1)}{6-2}=1$. The slope of PR is $\frac{1-(-1)}{0-2}=-1$. Since the product of slopes is $-1$, $PQ\perp PR$, making the triangle right-angled at P. For a right-angled triangle, the orthocentre is at the vertex of the right angle, which is $P(2,-1)$.