Question

Equation of the hyperbola whose vertices are (± 3, 0) and foci at (± 5, 0), is 

A. 16x² – 9y² = 144 

B. 9x² – 16y² = 144 

C. 25x² – 9y² = 225 

D. 9x² – 25y² = 81

Answer 1

Given: Vertices are (±3, 0) and foci are (±5, 0) 

To find: equation of the hyperbola 

Formula used: 

Standard form of the equation of hyperbola is,

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

Vertices of hyperbola are given by (±a, 0) 

Foci of hyperbola are given by (±ae, 0) 

Vertices are (±3, 0) and foci are (±5, 0) 

Therefore, 

a = 3 and ae = 5 

⇒ 3 × e = 5

\(\Rightarrow\) e = \(\frac{5}{3}\)

b² = a²(e² – 1)

⇒ b² = 16

Equation of hyperbola:

⇒ 16x² – 9y² = 144 

Hence, required equation of hyperbola is 16x² – 9y² = 144