The equation of the hyperbola with vertices (3, 0), (-3, 0) and semi-latus rectum 4 is given by:

Correct answer: C. 4x^ 2 - 3y^ 2 - 36 = 0 We have a = 3 ~and~ (b2)/(a) = 4 ⇒ b^ 2 = 12 Hence, the equation of the hyperbola is (x2)/(9) - (y2)/(12) = 1 ⇒ 4x^ 2 - 3y^ 2 = 36 ⇒ 4x^ 2 - 3y^ 2 - 36 = 0

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Hyperbola

JEE Main · Mathematics · Hyperbola

The equation of the hyperbola with vertices (3, 0), (-3, 0) and...

The equation of the hyperbola with vertices

\left(3, 0\right), \left(-3, 0\right)

and semi-latus rectum

4

is given by:

Options

A $4x^{2} - 3y^{2} + 36 = 0$
B $4x^{2} - 3y^{2} + 12 = 0$
C $4x^{2} - 3y^{2} - 36 = 0$
D $4x^{2} + 3y^{2} - 25 = 0$

Correct answer

C. $4x^{2} - 3y^{2} - 36 = 0$

Step-by-step solution

We have $a = 3$ ~and~ $\dfrac{b2}{a} = 4 ⇒ b^{2} = 12$
Hence, the equation of the hyperbola is
$\dfrac{x2}{9} - \dfrac{y2}{12} = 1$
$⇒ 4x^{2} - 3y^{2} = 36$
$⇒ 4x^{2} - 3y^{2} - 36 = 0$

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