Find the equation of the hyperbola whose vertices are (0, ± 3) and the eccentricity is 4/3 ?

CONCEPT:

The properties of a vertical hyperbola \(\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1\) are:

• Its centre is given by: (0, 0)
• Its foci are given by: (0, - ae) and (0, ae)
• Its vertices are given by: (0, - a)  and (0, a)
• Its eccentricity is given by: \(e = \frac{{\sqrt{{a^2} + {b^2}} }}{a}\)
• Length of transverse axis = 2a and its equation is x = 0.
• Length of conjugate axis = 2b and its equation is y = 0.
• Length of its latus rectum is given by: \(\frac{2b^2}{a}\)

CALCULATION:

Here, we have to find the equation of the hyperbola whose vertices are (0, ± 3) and the eccentricity is 4/3

By comparing the vertices (0, ± 3) with (0, ± a) we get

⇒ a = 3

As we know that, \(e = \frac{{\sqrt{{a^2} + {b^2}} }}{a}\)

⇒ a²e² = a² + b²

⇒ 16 = 9 + b²

⇒ b² = 7

so, the equation of the hyperbola is \(\frac{{{y^2}}}{{{9}}} - \frac{{{x^2}}}{{{7}}} = 1\)

Hence, option C is the correct answer.