the foci of the hyperbola 9x2 – 16y2 = 144 are

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the foci of the hyperbola 9x² – 16y² = 144 are

Given: 9x² – 16y² = 144

To find: coordinates of the foci f(m,n)

9x² – 16y² = 144

Formula used:

For hyperbola

Eccentricity(e) is given by,

Foci is given by (±ae, 0)

Here, a = 4 and b = 3

⇒ c = 5

Therefore,

Foci: (±5, 0)

The eccentricity of the hyperbola x² – 4y² = 1 is

The difference of the focal distances of any point on the hyperbola is equal to

The distance between the foci of a hyperbola is 16 and its eccentricity is , then equation of the hyperbola is

If e₁ is the eccentricity of the conic 9x² + 4y² = 36 and e₂ is the eccentricity of the conic 9x² – 4y² = 36, then