Question

find the focus and eccentricity of the eclipse 16x^2+9y^2=576.pls tell fast​

Answer 1

Answer:

Answer

We have, 16x2−9y2=576

⇒36x2−64y2=1

⇒62x2−82y2=1...(1)

On comparing equation (1) with the standard equation of hyperbola, a2x2−b2y2=1, 

we obtain a=6,b=8⇒e=1+a2b2=1+3664=35

Therefore, the coordinates of the foci are (±10,0)

The coordinates of the vertices are (±6,0)

Eccentricity e=35

Length of latus rectum = a2

Answer 2

Answer:

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