Let one focus of the hyperbola H: x^2/a^2 - y^2/b^2 = 1 be at (sqrt(10),0) and the corresponding directrix be x = 9/sqrt(10). If e and l respectively are the eccentricity and the length of the latus rectum of H, then 9(e^2 + l) is equal to :

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

Maths Q10 – JEE-MAIN 2025

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Home Maths Conic SectionsQuestion #130

Let one focus of the hyperbola H:

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

be at

(\sqrt{10}, 0)

and the corresponding directrix be

x = \frac{9}{\sqrt{10}}

. If

e

and

l

respectively are the eccentricity and the length of the latus rectum of H, then

9 (e^{2} + l)

is equal to :

Clue 1 of 3: Hyperbola Properties

Recall the standard equations for the focus and directrix of a hyperbola.

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● Video Walkthrough

Step 1: Determine 'a' and 'e' from Focus and Directrix✦ Active

For a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ , the focus is $(\pm a e, 0)$ and the corresponding directrix is $x = \pm \frac{a}{e}$ . Given the focus $(\sqrt{10}, 0)$ , we have $a e = \sqrt{10}$ . Given the directrix $x = \frac{9}{\sqrt{10}}$ , we have $\frac{a}{e} = \frac{9}{\sqrt{10}}$ .

(a e) \cdot (\frac{a}{e}) = \sqrt{10} \cdot \frac{9}{\sqrt{10}} ⟹ a^{2} = 9 ⟹ a = 3

Substitute $a = 3$ into $a e = \sqrt{10}$ : $3 e = \sqrt{10} ⟹ e = \frac{\sqrt{10}}{3}$ . Thus, $e^{2} = (\frac{\sqrt{10}}{3})^{2} = \frac{10}{9}$ .

Step 2: Calculate 'b^2' and the Latus Rectum 'l'○ Expand

The relationship between $a, b, e$ for a hyperbola is $b^{2} = a^{2} (e^{2} - 1)$ . Substitute the values of $a$ and $e^{2}$ :

b^{2} = 3^{2} (\frac{10}{9} - 1) = 9 (\frac{10 - 9}{9}) = 9 (\frac{1}{9}) = 1

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

l = \frac{2 (1)}{3} = \frac{2}{3}

Step 3: Compute the Final Expression○ Expand

We need to find the value of $9 (e^{2} + l)$ . Substitute the calculated values of $e^{2}$ and $l$ :

9 (e^{2} + l) = 9 (\frac{10}{9} + \frac{2}{3})

9 (e^{2} + l) = 9 (\frac{10}{9} + \frac{6}{9})

9 (e^{2} + l) = 9 (\frac{16}{9})

9 (e^{2} + l) = 16

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Maths Question 10 – JEE-MAIN 2025

Clue 1 of 3: Hyperbola Properties

🧠 Full Solution Path