Let the latus rectum of the hyperbola subtend an angle of at the centre of the hyperbola. If is equal to , where and m are co-prime numbers, then is equal to __________. [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let the latus rectum of the hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{b^{2}} = 1$ subtend an angle of $\frac{π}{3}$ at the centre of the hyperbola. If $b^{2}$ is equal to $\frac{l}{m} (1 + \sqrt{n})$ , where $l$ and m are co-prime numbers, then $l^{2} + m^{2} + n^{2}$ is equal to __________. [2024]

Ans.
(182)

In right triangle OFP, we have $\tan 30 ° = \frac{P F}{O F}$

$\Rightarrow \frac{1}{\sqrt{3}} = \frac{b^{2}}{a^{2} e} \Rightarrow e = \frac{\sqrt{3} b^{2}}{a^{2}}$

$e = \frac{\sqrt{3} b^{2}}{9}$ (Given $a^{2} = 9$ )

Squaring both sides, we get

$e^{2} = \frac{3 b^{4}}{81}$ ... (i)

and $e^{2} = 1 + \frac{b^{2}}{a^{2}} \Rightarrow e^{2} = 1 + \frac{b^{2}}{9}$ ... (ii)

Comparing equation (i) and (ii), we get

$\frac{3 b^{4}}{81} = 1 + \frac{b^{2}}{9} \Rightarrow b^{4} = 27 + 3 b^{2}$

$\Rightarrow b^{4} - 3 b^{2} - 27 = 0 \Rightarrow b^{2} = \frac{3}{2} (1 + \sqrt{13})$

According to given condition, $b^{2} = \frac{l}{m} (1 + \sqrt{n})$

On comparing, we get $l$ = 3, m = 2 and n = 13

$\Rightarrow l^{2} + m^{2} + n^{2} = {(3)}^{2} + {(2)}^{2} + {(13)}^{2} = 9 + 4 + 160 = 182$ .

Want Us to Email you About Special Offers & Updates?