Find the axes, eccentricity, latus rectum and the coordinates of the foci of the hyperbola `25 x^2-36 y^2=225.`

Answer

latus rectum `= 25/6`

coordinates of foci `=(±sqrt(61) / 2,0)`

Explanation

Equation of the hyperbola:

`25x2−36y^2=225`

rewriting the equation:

`((25x^2) /225)−( (36y^2) /225) = 1`


`(x^2 / 9) –(y^2/(225/36)) = 1`



Comparing with the equation of the hyperbola,

`x^2/a^2 – y^2/ b^2 = 1`


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