JEE Main 2026 — Parabola Question with Solution

The latus rectum of the parabola $P:y^2=4kx$ is the line segment $x=k$. The endpoints of this latus rectum are $(k,2k)$ and $(k,-2k)$.

The latus rectum of the ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is the line segment $x=ae$ (taking the one in the positive $x$-axis). The endpoints of this latus rectum are $\left(ae,\frac{b^2}{a}\right)$ and $\left(ae,-\frac{b^2}{a}\right)$.

Since the line segment joining the points of intersection of $P$ and $E$ is the latus rectum for both curves, their endpoints must coincide. Comparing the coordinates, we get:
$k=ae$
$2k=\frac{b^2}{a}$

Substituting $k=ae$ into the second equation:
$2ae=\frac{b^2}{a}$
$2a^{2}e=b^2$

For an ellipse, the relation between $a,b,$ and $e$ is $b^2=a^2(1-e^2)$. Substituting this into the equation above:
$2a^{2}e=a^2(1-e^2)$

Since $a\ne 0$, we can divide by $a^2$:
$2e=1-e^2$
$e^2+2e-1=0$

Solving for $e$ using the quadratic formula:
$e=\frac{-2\pm \sqrt{4-4(1)(-1)}}{2}=\frac{-2\pm 2\sqrt{2}}{2}=-1\pm \sqrt{2}$

Since the eccentricity of an ellipse must satisfy $0<e<1$, we take the positive root:
$e=\sqrt{2}-1$

We need to find the value of $e^2+2\sqrt{2}$. First, calculate $e^2$:
$e^2=(\sqrt{2}-1{)}^2=2-2\sqrt{2}+1=3-2\sqrt{2}$

Therefore:
$e^2+2\sqrt{2}=(3-2\sqrt{2})+2\sqrt{2}=3$

Answer: $3$