JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Parabola

For the parabola $y^2=36x$, we have $4a=36$, so $a=9$. A focal chord has endpoints $P(9t^2,18t)$ and $Q(9/t^2,-18/t)$. The length is $9(t+1/t{)}^2=100$, giving $t+1/t=10/3$. Since the ordinate of P is positive and the chord makes an acute angle with the positive x-axis, $t=3$. Thus $P(81,54)$ and $Q(1,-6)$. Using the section formula with $PM:MQ=3:1$, we get $M(21,9)$. The slope of PQ is $(54-(-6))/(81-1)=60/80=3/4$. The perpendicular line through M has slope $-4/3$, giving equation $y-9=-\frac{4}{3}(x-21)$, or $4x+3y=111$. Checking each point: $(-3,43)$ gives $4(-3)+3(43)=-12+129=117\ne 111$, so it does NOT lie on the line.