JEE Main 2026 — Mathematics Coordinate Geometry

The latus rectum of the parabola $P: y^2 = 4kx$P:y2=4kx is the line segment $x = k$x=k. The endpoints of this latus rectum are $(k, 2k)$(k,2k) and $(k, -2k)$(k,−2k).

The latus rectum of the ellipse $E: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$E:a2x2​+b2y2​=1 is the line segment $x = ae$x=ae (taking the one in the positive $x$x-axis). The endpoints of this latus rectum are $\left(ae, \dfrac{b^2}{a}\right)$(ae,ab2​) and $\left(ae, -\dfrac{b^2}{a}\right)$(ae,−ab2​).

Since the line segment joining the points of intersection of $P$P and $E$E is the latus rectum for both curves, their endpoints must coincide. Comparing the coordinates, we get:

$$k = ae$$k=ae

$$2k = \dfrac{b^2}{a}$$2k=ab2​

Substituting $k = ae$k=ae into the second equation:

$$2ae = \dfrac{b^2}{a}$$2ae=ab2​

$$2a^2e = b^2$$2a2e=b2

For an ellipse, the relation between $a, b,$a,b, and $e$e is $b^2 = a^2(1 - e^2)$b2=a2(1−e2). Substituting this into the equation above:

$$2a^2e = a^2(1 - e^2)$$2a2e=a2(1−e2)

Since $a \neq 0$a=0, we can divide by $a^2$a2:

$$2e = 1 - e^2$$2e=1−e2

$$e^2 + 2e - 1 = 0$$e2+2e−1=0

Solving for $e$e using the quadratic formula:

$$e = \dfrac{-2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \dfrac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$$e=2−2±4−4(1)(−1)​​=2−2±22​​=−1±2​

Since the eccentricity of an ellipse must satisfy $0 < e < 1$0<e<1, we take the positive root:

$$e = \sqrt{2} - 1$$e=2​−1

We need to find the value of $e^2 + 2\sqrt{2}$e2+22​. First, calculate $e^2$e2:

$$e^2 = (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}$$e2=(2​−1)2=2−22​+1=3−22​

Therefore:

$$e^2 + 2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$$e2+22​=(3−22​)+22​=3

Answer: $3$3