Maths Q11 – JEE-MAIN 2026

The eccentricity of an ellipse E with centre at the origin is

\frac{\sqrt{3}}{2}

and its directrices are

x = \pm \frac{4 \sqrt{6}}{3}

. Let H:

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose length of latus rectum is equal to the length of minor axis of E. Then the distance between the foci of H is:

Clue 1 of 3: Understand the properties of an ellipse and a hyperbola

Recall the definitions of eccentricity, directrices, semi-major axis, semi-minor axis, latus rectum, and foci for both conic sections.

🧠 Full Solution Path

Step 1: Determine parameters of Ellipse E✦ Active

Given the eccentricity of ellipse E is $e_{E} = \frac{\sqrt{3}}{2}$ and its directrices are $x = \pm \frac{4 \sqrt{6}}{3}$ . The directrix equation for an ellipse is $x = \pm \frac{a_{E}}{e_{E}}$ . Thus, we have:

\frac{a_{E}}{e_{E}} = \frac{4 \sqrt{6}}{3} ⟹ a_{E} = e_{E} \cdot \frac{4 \sqrt{6}}{3} = \frac{\sqrt{3}}{2} \cdot \frac{4 \sqrt{6}}{3} = \frac{4 \sqrt{18}}{6} = \frac{4 \cdot 3 \sqrt{2}}{6} = 2 \sqrt{2}

Now, find the semi-minor axis $b_{E}$ using the relation $b_{E}^{2} = a_{E}^{2} (1 - e_{E}^{2})$ :

b_{E}^{2} = (2 \sqrt{2})^{2} (1 - {(\frac{\sqrt{3}}{2})}^{2}) = 8 (1 - \frac{3}{4}) = 8 (\frac{1}{4}) = 2

So, $b_{E} = \sqrt{2}$ . The length of the minor axis of E is $2 b_{E} = 2 \sqrt{2}$ .

Step 2: Determine parameters of Hyperbola H○ Expand

For hyperbola H, its eccentricity $e_{H}$ is equal to the length of the semi-major axis of E, and its latus rectum $L R_{H}$ is equal to the length of the minor axis of E. So:

e_{H} = a_{E} = 2 \sqrt{2}

The length of the latus rectum of H is $L R_{H} = \frac{2 b_{H}^{2}}{a_{H}}$ , which is equal to $2 b_{E}$ :

\frac{2 b_{H}^{2}}{a_{H}} = 2 \sqrt{2} ⟹ b_{H}^{2} = \sqrt{2} a_{H}

For a hyperbola, the relation between $a_{H}$ , $b_{H}$ , and $e_{H}$ is $b_{H}^{2} = a_{H}^{2} (e_{H}^{2} - 1)$ . Substitute $e_{H} = 2 \sqrt{2}$ :

b_{H}^{2} = a_{H}^{2} ((2 \sqrt{2})^{2} - 1) = a_{H}^{2} (8 - 1) = 7 a_{H}^{2}

Equating the two expressions for $b_{H}^{2}$ :

7 a_{H}^{2} = \sqrt{2} a_{H}

Since $a_{H} \neq 0$ , we can divide by $a_{H}$ :

7 a_{H} = \sqrt{2} ⟹ a_{H} = \frac{\sqrt{2}}{7}

💡 Teacher's Secret Hint

Be careful to distinguish between 'semi-minor axis' ( $b_{E}$ ) and 'minor axis' ( $2 b_{E}$ ) when setting up the latus rectum condition for the hyperbola.

Step 3: Calculate the distance between foci of H○ Expand

The distance between the foci of a hyperbola is $2 c_{H} = 2 a_{H} e_{H}$ . Substitute the values of $a_{H}$ and $e_{H}$ :

2 c_{H} = 2 \cdot (\frac{\sqrt{2}}{7}) \cdot (2 \sqrt{2})

2 c_{H} = \frac{4 \cdot (\sqrt{2})^{2}}{7} = \frac{4 \cdot 2}{7} = \frac{8}{7}

The distance between the foci of H is $\frac{8}{7}$ .

💡 Teacher's Secret Hint

Ensure all calculations involving square roots and fractions are simplified correctly.

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Maths Question 11 – JEE-MAIN 2026

Clue 1 of 3: Understand the properties of an ellipse and a hyperbola

🧠 Full Solution Path