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Hyperbola

Section: Mathematics

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Question : 15 of 34

Marks: +1, -0

One of the latus recta of the hyperbola

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

subtends an angle

2 \tan^{-1} \left( \frac{3}{2} \right)

at the centre of the hyperbola. If

b^2 = 36

and e is the eccentricity of the given hyperbola, then

\sqrt{a^2 + e^2} =

[AP EAPCET 21 May 2025 S1]

$4$
$\sqrt{14}$
$6$
$\sqrt{21}$

Solution:

Given,

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

and

b^2 = 36

Coordinate of one of

e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{36}{a^2}

The latus rectum is

L \left(a e, + \frac{b^2}{a}\right)

and

L' \left(a e, - \frac{b^2}{a}\right)

.

Now, slope of

O L = \frac{b^2}{a^2 e}

Slope of

O L' = - \frac{b^2}{a^2 e}

Now, angle between

O L

and

O L'

2 \tan^{-1} \frac{b^2}{a^2 e} = 2 \tan^{-1} \frac{3}{2}

\therefore \;\; \frac{b^2}{a^2+e} = \frac{3}{2} \Rightarrow \frac{3e}{2} = \frac{b^2}{a^2}

\frac{3e}{2} = e^2 - 1 \qquad \left[ \because e^2 = 1 + \frac{b^2}{a^2} \right]

2e^2 - 3e - 2 = 0

(e-2)(2e+1) = 10 \Rightarrow e=2, e \neq -\frac{1}{2}

a^2 = \frac{2b^2}{3e} = \frac{2 \times 36}{3 \times 2} = 12 \qquad [ \because b^2 = 36 ]

\therefore \sqrt{a^2+e^2} = \sqrt{12+4} = 4

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