JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Parabola

The circle has centre $S(2,8)$ and radius 2. The normal to the parabola $y^2=4x$ at $(t^2,2t)$ is $y+tx=2t+t^3$. For this to pass through $S(2,8)$: $8+2t=2t+t^3$, giving $t=2$. So $P=(4,4)$. Then $SP=\sqrt{(4-2{)}^2+(4-8{)}^2}=\sqrt{4+16}=2\sqrt{5}$, confirming (A). The normal at $P$ is $y+2x=12$, which has $x$-intercept 6, confirming (C). Point $Q$ divides $SP$ such that $SQ=2$ (radius), so $QP=2\sqrt{5}-2$. The tangent to the circle at $Q$ is perpendicular to $SQ$. The slope of $SP$ is $\frac{4-8}{4-2}=-2$, so the slope of the tangent at $Q$ is $\frac{1}{2}$, confirming (D). Option (B) is incorrect as the ratio $SQ:QP=2:(2\sqrt{5}-2)=1:(\sqrt{5}-1)$, not $(\sqrt{5}+1):2$.