Maths Q13 – JEE-MAIN 2026

Let the eccentricity

e

of a hyperbola satisfy the equation

6 e^{2} - 11 e + 3 = 0

. If the foci of the hyperbola are

(3, 5)

and

(3, - 4)

, then the length of its latus rectum is :

Clue 1 of 3: Eccentricity of Hyperbola

First, solve the quadratic equation to find the eccentricity $e$ . Remember that for a hyperbola, $e > 1$ .

🧠 Full Solution Path

Step 1: Determine Eccentricity and Foci Properties✦ Active

Solve the quadratic equation $6 e^{2} - 11 e + 3 = 0$ for $e$ . Factoring or using the quadratic formula yields $e = \frac{11 \pm \sqrt{121 - 72}}{12} = \frac{11 \pm 7}{12}$ . The possible values are $e = \frac{18}{12} = \frac{3}{2}$ and $e = \frac{4}{12} = \frac{1}{3}$ . For a hyperbola, the eccentricity $e$ must be greater than 1, so $e = \frac{3}{2}$ .

The foci are given as $(3, 5)$ and $(3, - 4)$ . The distance between the foci, $2 c$ , is the distance between these two points. $2 c = \sqrt{(3 - 3)^{2} + (5 - (- 4))^{2}} = \sqrt{0^{2} + 9^{2}} = 9$ . Therefore, $c = \frac{9}{2}$ .

Step 2: Calculate 'a' and 'b²'○ Expand

Using the relationship between eccentricity, focal distance, and semi-transverse axis length, $e = \frac{c}{a}$ , we can find $a$ . Substitute the values: $\frac{3}{2} = \frac{9 / 2}{a}$ . Solving for $a$ gives $a = \frac{9 / 2}{3 / 2} = 3$ .

For a hyperbola, the relationship between $a$ , $b$ , and $c$ is $c^{2} = a^{2} + b^{2}$ . Substitute the values of $c$ and $a$ : $(\frac{9}{2})^{2} = (3)^{2} + b^{2}$ . This simplifies to $\frac{81}{4} = 9 + b^{2}$ . Solving for $b^{2}$ : $b^{2} = \frac{81}{4} - 9 = \frac{81 - 36}{4} = \frac{45}{4}$ .

Step 3: Find the Length of the Latus Rectum○ Expand

The length of the latus rectum of a hyperbola is given by the formula $\frac{2 b^{2}}{a}$ . Substitute the calculated values of $b^{2}$ and $a$ :

Length of Latus Rectum = \frac{2 \times \frac{45}{4}}{3} = \frac{\frac{45}{2}}{3} = \frac{45}{2 \times 3} = \frac{45}{6} = \frac{15}{2}

Thus, the length of the latus rectum is $\frac{15}{2}$ .

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Maths Question 13 – JEE-MAIN 2026

Clue 1 of 3: Eccentricity of Hyperbola

🧠 Full Solution Path