Let H: x^2/a^2 - y^2/b^2 = 1 be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is 8/3. If the line x = alpha intersects the hyperbola H at the points A and B such that the area of the triangle AOB is 4sqrt(15), where O is the origin, then a^2 equals

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q11 – JEE-MAIN 2026

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Home Maths Conic SectionsQuestion #113

Let H:

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

be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is

\frac{8}{3}

. If the line

x = α

intersects the hyperbola H at the points A and B such that the area of the triangle AOB is

4 \sqrt{15}

, where O is the origin, then

a^{2}

equals

Clue 1 of 3: Hyperbola Properties

Recall the definitions for the distance between foci ( $2 a e$ ) and directrices ( $2 a / e$ ) for a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ .

🧠 Full Solution Path

Step 1: Determine Hyperbola Parameters✦ Active

For the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ : The distance between foci is $2 a e = 6 ⟹ a e = 3$ . The distance between directrices is $\frac{2 a}{e} = \frac{8}{3} ⟹ \frac{a}{e} = \frac{4}{3}$ . Multiplying these two equations gives $(a e) (\frac{a}{e}) = 3 \times \frac{4}{3} ⟹ a^{2} = 4$ . Dividing them gives $\frac{a e}{a / e} = \frac{3}{4 / 3} ⟹ e^{2} = \frac{9}{4}$ . Using the relation $b^{2} = a^{2} (e^{2} - 1)$ , we find $b^{2} = 4 (\frac{9}{4} - 1) = 4 (\frac{5}{4}) = 5$ $. T h u s, t h e h y p e r b o l a e q u a t i o n i s$ $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$ $.

Step 2: Find Intersection Points and Triangle Area○ Expand

The line $x = α$ intersects the hyperbola. Substitute $x = α$ into the hyperbola equation: $\frac{α^{2}}{4} - \frac{y^{2}}{5} = 1 ⟹ \frac{y^{2}}{5} = \frac{α^{2}}{4} - 1 = \frac{α^{2} - 4}{4}$ . This gives $y^{2} = \frac{5 (α^{2} - 4)}{4}$ , so $y = \pm \frac{\sqrt{5}}{2} \sqrt{α^{2} - 4}$ $. F o r t w o d i s t i n c t p o i n t s A a n d B, w e m u s t h a v e$ \alpha^2 - 4 > 0 $, i . e .,$ |\alpha| > 2 $. T h e p o i n t s a r e A$ (\alpha, \frac{\sqrt{5}}{2} \sqrt{\alpha^2 - 4}) $a n d B$ (\alpha, -\frac{\sqrt{5}}{2} \sqrt{\alpha^2 - 4}) $. T h e b a s e A B h a s l e n g t h$ $\sqrt{5} \sqrt{α^{2} - 4}$ $a n d t h e h e i g h t o f t h e t r i a n g l e f r o m t h e o r i g i n O t o t h e l i n e$ x=\alpha $i s$ |\alpha| $. T h e a r e a o f$ \triangle AOB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{5} \sqrt{\alpha^2 - 4} \times |\alpha|$$.

Step 3: Solve for

α^{2}

○ Expand

Given that the area of $△ A O B$ is $4 \sqrt{15}$ : $\frac{1}{2} \times \sqrt{5} \sqrt{α^{2} - 4} \times | α | = 4 \sqrt{15}$ $\sqrt{5} \sqrt{α^{2} - 4} \times | α | = 8 \sqrt{15}$ Divide by $\sqrt{5}$ : $\sqrt{α^{2} - 4} \times | α | = 8 \sqrt{3}$ Square both sides: $(α^{2} - 4) α^{2} = (8 \sqrt{3})^{2} = 64 \times 3 = 192$ Let $X = α^{2}$ . The equation becomes $X^{2} - 4 X - 192 = 0$ . Factoring the quadratic equation: $(X - 16) (X + 12) = 0$ . This yields $X = 16$ or $X = - 12$ . Since $X = α^{2}$ must be positive, we take $X = 16$ . Thus, $α^{2} = 16$ . This value satisfies the condition $| α | > 2$ (since $| α | = 4$ ). Although the question asks for $a^{2}$ , the value of $a^{2}$ from the hyperbola definition is 4, which is not among the options. Given the context of competitive exams, it is highly probable that the question implicitly asks for the value of $α^{2}$ .

💡 Teacher's Secret Hint

Always check if the calculated value satisfies any initial conditions or constraints, such as $| α | > 2$ in this case.

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Maths Question 11 – JEE-MAIN 2026

Clue 1 of 3: Hyperbola Properties

🧠 Full Solution Path