JEE Main 2026 — Mathematics Coordinate Geometry

For the ellipse $E$E, the eccentricity is $e_E = \dfrac{\sqrt{3}}{2}$eE​=23​​ and the directrices are $x = \pm \dfrac{a_E}{e_E} = \pm \dfrac{4\sqrt{6}}{3}$x=±eE​aE​​=±346​​.

The semi-major axis $a_E$aE​ is given by:

$$a_E = e_E \times \dfrac{4\sqrt{6}}{3} = \dfrac{\sqrt{3}}{2} \times \dfrac{4\sqrt{6}}{3} = \dfrac{12\sqrt{2}}{6} = 2\sqrt{2}$$aE​=eE​×346​​=23​​×346​​=6122​​=22​

The semi-minor axis $b_E$bE​ is given by:

$$b_E^2 = a_E^2(1 - e_E^2) = (2\sqrt{2})^2 \left(1 - \dfrac{3}{4}\right) = 8 \times \dfrac{1}{4} = 2 \Rightarrow b_E = \sqrt{2}$$bE2​=aE2​(1−eE2​)=(22​)2(1−43​)=8×41​=2⇒bE​=2​

For the hyperbola $H: \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$H:a2x2​−b2y2​=1, its eccentricity $e_H$eH​ is equal to the semi-major axis of $E$E:

$$e_H = a_E = 2\sqrt{2}$$eH​=aE​=22​

The length of the latus rectum of $H$H is equal to the length of the minor axis of $E$E ($2b_E$2bE​):

$$\dfrac{2b^2}{a} = 2b_E = 2\sqrt{2} \Rightarrow b^2 = \sqrt{2}a$$a2b2​=2bE​=22​⇒b2=2​a

Using the standard relation for a hyperbola $b^2 = a^2(e_H^2 - 1)$b2=a2(eH2​−1), we substitute $b^2$b2 and $e_H$eH​:

$$\sqrt{2}a = a^2((2\sqrt{2})^2 - 1)$$2​a=a2((22​)2−1)

$$\sqrt{2}a = a^2(8 - 1) = 7a^2$$2​a=a2(8−1)=7a2

Since $a > 0$a>0, dividing by $a$a gives:

$$a = \dfrac{\sqrt{2}}{7}$$a=72​​

The distance between the foci of the hyperbola $H$H is $2ae_H$2aeH​:

$$2ae_H = 2 \times \dfrac{\sqrt{2}}{7} \times 2\sqrt{2} = \dfrac{8}{7}$$2aeH​=2×72​​×22​=78​

Answer: $\dfrac{8}{7}$78​