# Find the equation of hyperbola whose eccentricity is 5/3 and whose vertices are (0, ± 6) ?

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Find the equation of hyperbola whose eccentricity is 5/3 and whose vertices are (0, ± 6) ?
1. $\frac{{{y^2}}}{{{36}}} - \frac{{{x^2}}}{{{81}}} = 1$
2. $\frac{{{y^2}}}{{{36}}} - \frac{{{x^2}}}{{{49}}} = 1$
3. $\frac{{{y^2}}}{{{36}}} - \frac{{{x^2}}}{{{64}}} = 1$
4. $\frac{{{y^2}}}{{{25}}} - \frac{{{x^2}}}{{{64}}} = 1$
5. None of these

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Correct Answer - Option 3 : $\frac{{{y^2}}}{{{36}}} - \frac{{{x^2}}}{{{64}}} = 1$

Concept:

The hyperbola of the form $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$ has:

• Centre is given by: (0, 0)
• Vertices are given by: (0, ± a)
• Foci are given by: (0, ± c)
• Length of transverse axis is given by: 2a
• Length of conjugate axis is given by: 2b
• Eccentricity is given by:$e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}$
• b2 = c2 - a2

Calculation:

Given: The vertices and eccentricity of hyperbola are: (0, ± 6) and 5/3 respectively.

⇒ e = 5/3

∵ The vertices of the given hyperbola are of the form (0, ± a), it is a vertical hyperbola i.e it is of the form: $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$

As we that the vertices of the vertical parabola i.e $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$ are of the form (0, ± a).

⇒ a = 6 and a2 = 36

As we know that eccentricity e = c/a

⇒ c = e ⋅ a

By substituting e = 5/3 and a = 6 in the above equation we get

⇒ c = 10 and c2 = 100

As we know that, b2 = c2 - a2

⇒ b2 =  100 - 36 = 64

Hence the equation of required hyperbola is: $\frac{{{y^2}}}{{{36}}} - \frac{{{x^2}}}{{{64}}} = 1$