Find the centre, eccentricity, foci and directrices of the hyperbola : `16 x^2-9y^2+32 x+36 y-164=0`

Answer

Eccentricity of the hyperbola =`5/3`

Coordinates of Foci ` =(−6,2),(4,2)`


Equation of directrices `5x−4=0 or 5x+14=0`

Explanation

ATQ, the equation `16x^2 − 9y^2 + 32x + 36y – 164 = 0` can be re-written by:


`⇒16(x^2+2x)−9(y^2−4y)=164`



`⇒16(x^2+2x+1)−9(y^2−4y+4)=164+16−36`



`⇒16(x+1)^2−9(y−2)^2=144`



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