If a directrix of a hyperbola centered at the origin and passing through the point (4, - 2 3 ) is 5x = 4 5 ~and~its eccentricity is e , then:

Correct answer: B. 4e^ 4 - 24e^ 2 + 35 = 0 Let equation of hyperbola is (x2)/(a2) - (y2)/(b2) = 1 ∴ It passes through (4, - 2 3 ) ⇒ (16)/(a2) - (12)/(b2) = 1 ⇒ 16 - 12 × (a2)/(b2) = a^ 2 . . . (1) Equation of directrix is x = (a)/(e) = (4)/(5),given in question. ⇒ a^ 2 = (16)/(5) e^ 2 . . . (2) And we know that b^ 2 = a^ 2 (e^ 2 - 1) ⇒ (b2)/(a2) = e^ 2 - 1 . . . ( 3) ∴ From~ ( 1), ( 2) and (3) 16 - 12 e^ 2 - 1 = (16)/(5) e^ 2 ⇒ 16e^ 2 - 16 - 12 = 16e^ 2 5 (e^ 2 - 1) ⇒ 4e^ 4 - 24e^ 2 + 35 = 0

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Hyperbola

JEE Main · Mathematics · Hyperbola

If a directrix of a hyperbola centered at the origin and passing...

If a directrix of a hyperbola centered at the origin and passing through the point

\left(4, - 2\sqrt{3}\right)

5x = 4\sqrt{5}

~and~its eccentricity is

e

, then:

Options

A $4e^{4} + 8e^{2} - 35 = 0$
B $4e^{4} - 24e^{2} + 35 = 0$
C $4e^{4} - 24e^{2} + 27 = 0$
D $4e^{4} - 12e^{2} - 27 = 0$

Correct answer

B. $4e^{4} - 24e^{2} + 35 = 0$

Step-by-step solution

Let equation of hyperbola is $\dfrac{x2}{a2} - \dfrac{y2}{b2} = 1$

$∴$ It passes through \left(4, - 2\sqrt{3}\right) ⇒ \dfrac{16}{a2} - \dfrac{12}{b2} = 1 $$ ⇒ 16 - 12 \times \dfrac{a2}{b2} = a^{2} . . . \left(1\right) $Equation of directrix is$ x = $\dfrac{a}{e} = \dfrac{4}{5},given in question.$ ⇒ a^{2} = $\dfrac{16}{5}$ e^{2} . . . \left(2\right) $And we know that$ b^{2} = a^{2} \left(e^{2} - 1\right) $$ ⇒ $\dfrac{b2}{a2} =$ e^{2} - 1 . . . \left( $3\right)$

$∴$ From~ \left( $1\right), \left($ 2\right) \text{ and } \left(3\right) $$ 16 - $\dfrac{12}{e^{2} - 1} = \dfrac{16}{5}$ e^{2} $$ ⇒ 16e^{2} - 16 - 12 = $\dfrac{16e^{2}}{5} \left(e^{2} - 1\right)$

$⇒ 4e^{4} - 24e^{2} + 35 = 0$

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