Write the coordinates of the foci of the hyperbola `9x^2-16 y^2=144.`

Explanation

We have giventhe equation of hyperbola as `9x^2-16 y^2=144.`

`(9x^2)/(12)^2− (16y^2)/(12)^2=1`

`x^2/16− y^2/9=1`

Now compare with standard equation `x^2/a^2− y^2/b^2=1`
we have

`a^2=16, b^2=9`

Therefore, `e^2=1+(b^2/a^2)`

`e^2=1+(9/16)`

`e^2=(16+9)/16`

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