JEE Advanced 2024 · Paper 2, Mathematics question 7

For $y^2=8x$ we have $a=2$. The tangent at $(2t^2,4t)$ is $ty=x+2t^2$.

It passes through $C_1(-4,0)$ when $0=-4+2t^2$, i.e. $t=\pm\sqrt2$. So
$$A_1=(4,4\sqrt2),\qquad B_1=(4,-4\sqrt2).$$

• (A) $OA_1=\sqrt{16+32}=4\sqrt3$. ✓

• (B) $A_1B_1=8\sqrt2\ne16$. ✗

• (C)/(D) The triangle is symmetric about the $x$-axis, so its orthocentre lies on $y=0$.

  • The slope of $B_1C_1$ is $\dfrac{-4\sqrt2}{8}=-\dfrac{1}{\sqrt2}$, so the altitude from $A_1$ has slope $\sqrt2$: $y-4\sqrt2=\sqrt2(x-4)$.
  • At $y=0$ this gives $x=0$, so the orthocentre is $(0,0)$.

  (C) is true and (D) is false.