JEE Main 2026 — Parabola Question with Solution

The equation of the parabola is $y^2=16x$, which gives $a=4$. Any point on the parabola can be taken as $(4t^2,8t)$.

The vertex $B$ is given as $(4,8)$, which corresponds to the parameter $t=1$. Let the parameters for vertices $A$ and $C$ be $t_1$ and $t_2$ respectively. Thus, $A\equiv (4t_1^2,8t_1)$ and $C\equiv (4t_2^2,8t_2)$.

The slope of $AB$ is:
$m_{AB}=\frac{8t_1-8}{4t_1^2-4}=\frac{2(t_1-1)}{(t_1-1)(t_1+1)}=\frac{2}{t_1+1}$

Similarly, the slope of $BC$ is:
$m_{BC}=\frac{2}{t_2+1}$

Since $△ABC$ is right-angled at $B$, $AB\perp BC$. Therefore, $m_{AB}\times m_{BC}=-1$:
$\frac{2}{t_1+1}\times \frac{2}{t_2+1}=-1$
$(t_1+1)(t_2+1)=-4$
$t_1{t}_2+t_1+t_2+5=0$

Let the centroid of $△ABC$ be $(h,k)$. Then:
$h=\frac{4+4t_1^2+4t_2^2}{3}\Rightarrow t_1^2+t_2^2=\frac{3h}{4}-1$
$k=\frac{8+8t_1+8t_2}{3}\Rightarrow t_1+t_2=\frac{3k}{8}-1$

From the perpendicularity condition, we can express $t_1{t}_2$ as:
$t_1{t}_2=-5-(t_1+t_2)=-5-\left(\frac{3k}{8}-1\right)=-4-\frac{3k}{8}$

Using the algebraic identity $t_1^2+t_2^2=(t_1+t_2{)}^2-2t_1{t}_2$, we substitute the expressions in terms of $k$:
$\frac{3h}{4}-1={\left(\frac{3k}{8}-1\right)}^2-2\left(-4-\frac{3k}{8}\right)$
$\frac{3h}{4}-1=\frac{9k^2}{64}-\frac{3k}{4}+1+8+\frac{3k}{4}$
$\frac{3h}{4}-1=\frac{9k^2}{64}+9$
$\frac{3h}{4}=\frac{9k^2}{64}+10$
$3h=\frac{9k^2}{16}+40$
$\frac{9k^2}{16}=3h-40$
$k^2=\frac{16}{9}(3h-40)=\frac{16}{3}\left(h-\frac{40}{3}\right)$

Replacing $(h,k)$ with $(x,y)$, the locus $C_o$ is the parabola:
$y^2=\frac{16}{3}\left(x-\frac{40}{3}\right)$

The length of the latus rectum of this parabola is $\frac{16}{3}$.

Three times the length of the latus rectum is $3\times \frac{16}{3}=16$.

Answer: $16$