If the eccentricity e of the hyperbola x^2/a^2 - y^2/b^2 = 1, passing through (6,4√3), satisfies 15(e^2 + 1) = 34e, then the length of the latus rectum of the hyperbola x^2/b^2 - y^2/2(a^2+1) = 1 is:

The correct answer is Option 1. 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 #124

If the eccentricity

e

of the hyperbola

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

, passing through

(6, 4 \sqrt{3})

, satisfies

15 (e^{2} + 1) = 34 e

, then the length of the latus rectum of the hyperbola

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

is:

Clue 1 of 3: Hyperbola Properties

Recall the relationship between $a$ , $b$ , and eccentricity $e$ for a hyperbola, and the formula for the length of the latus rectum.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Determine the Eccentricity (

e

)✦ Active

The given equation for eccentricity is $15 (e^{2} + 1) = 34 e$ . Rearrange it into a standard quadratic form:

15 e^{2} - 34 e + 15 = 0

Solve this quadratic equation for $e$ using the quadratic formula $e = \frac{- B \pm \sqrt{B^{2} - 4 A C}}{2 A}$ :

e = \frac{34 \pm \sqrt{(- 34)^{2} - 4 (15) (15)}}{2 (15)} = \frac{34 \pm \sqrt{1156 - 900}}{30} = \frac{34 \pm \sqrt{256}}{30} = \frac{34 \pm 16}{30}

This gives two possible values for $e$ : $e_{1} = \frac{34 + 16}{30} = \frac{50}{30} = \frac{5}{3}$ and $e_{2} = \frac{34 - 16}{30} = \frac{18}{30} = \frac{3}{5}$ . Since the eccentricity of a hyperbola must be greater than 1 ( $e > 1$ ), we choose $e = \frac{5}{3}$ .

Step 2: Find

a^{2}

and

b^{2}

○ Expand

The hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ passes through $(6, 4 \sqrt{3})$ . Substitute these coordinates:

\frac{6^{2}}{a^{2}} - \frac{(4 \sqrt{3})^{2}}{b^{2}} = 1 ⟹ \frac{36}{a^{2}} - \frac{48}{b^{2}} = 1 (1)

The relationship between $a$ , $b$ , and $e$ for a hyperbola is $b^{2} = a^{2} (e^{2} - 1)$ . Substitute $e = \frac{5}{3}$ :

b^{2} = a^{2} ({(\frac{5}{3})}^{2} - 1) = a^{2} (\frac{25}{9} - 1) = a^{2} (\frac{16}{9}) (2)

Substitute (2) into (1):

\frac{36}{a^{2}} - \frac{48}{\frac{16 a^{2}}{9}} = 1 ⟹ \frac{36}{a^{2}} - \frac{48 \cdot 9}{16 a^{2}} = 1 ⟹ \frac{36}{a^{2}} - \frac{27}{a^{2}} = 1

This simplifies to $\frac{9}{a^{2}} = 1$ , so $a^{2} = 9$ . Now, substitute $a^{2} = 9$ back into (2) to find $b^{2}$ :

b^{2} = \frac{16}{9} (9) = 16

Thus, $a^{2} = 9$ and $b^{2} = 16$ (which implies $b = 4$ since $b$ is a length).

💡 Teacher's Secret Hint

Ensure to use the correct eccentricity value ( $e > 1$ ) for a hyperbola.

Step 3: Calculate the Latus Rectum Length○ Expand

We need to find the length of the latus rectum for the hyperbola $\frac{x^{2}}{b^{2}} - \frac{y^{2}}{2 (a^{2} + 1)} = 1$ . This is of the form $\frac{x^{2}}{A^{2}} - \frac{y^{2}}{B^{2}} = 1$ , where $A^{2} = b^{2}$ and $B^{2} = 2 (a^{2} + 1)$ . The length of the latus rectum is given by $\frac{2 B^{2}}{A}$ .

Length of Latus Rectum = \frac{2 \cdot [2 (a^{2} + 1)]}{\sqrt{b^{2}}} = \frac{4 (a^{2} + 1)}{b}

Substitute the values $a^{2} = 9$ and $b = 4$ :

Length of Latus Rectum = \frac{4 (9 + 1)}{4} = \frac{4 (10)}{4} = 10

💡 Teacher's Secret Hint

Carefully identify $A^{2}$ and $B^{2}$ for the second hyperbola before applying the latus rectum formula.

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Maths Question 11 – JEE-MAIN 2026

Clue 1 of 3: Hyperbola Properties

🧠 Full Solution Path