Q. 15.0( 1 Vote )

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

A. 16x^{2} – 9y^{2} = 144

B. 9x^{2} – 16y^{2} = 144

C. 25x^{2} – 9y^{2} = 225

D. 9x^{2} – 25y^{2} = 81

Answer :

**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,

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

b^{2} = a^{2}(e^{2} – 1)

⇒ b^{2} = 16

Equation of hyperbola:

⇒ 16x^{2} – 9y^{2} = 144

Hence, required equation of hyperbola is **16x ^{2} – 9y^{2} = 144**

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