What is the vertices 16x^2+9y^2-96x+36y-36=0 ?

The vertices 16x^2+9y^2-96x+36y-36=0 is (3,-2+2sqrt(6)),(3,-2-2sqrt(6))

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vertices 16x^2+9y^2-96x+36y-36=0

vertices

16 x^{2} + 9 y^{2} - 96 x + 36 y - 36 = 0

(3, - 2 + 26), (3, - 2 - 26)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

16 x^{2} + 9 y^{2} - 96 x + 36 y - 36 = 0 :

Ellipse with center

(h, k) = (3, - 2), b = 26, a = \frac{3 3}{2}

(3, - 2 + 26), (3, - 2 - 26)

\frac{x ^{2}}{7 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

25 x^{2} + 9 y^{2} - 250 x + 72 y + 544 = 0

x^{2} + 16 y^{2} = 16

\frac{( x + 3 ) ^{2}}{4} + \frac{( y - 2 ) ^{2}}{16} = 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{9} = 1

What is the vertices 16x^2+9y^2-96x+36y-36=0 ?
The vertices 16x^2+9y^2-96x+36y-36=0 is (3,-2+2sqrt(6)),(3,-2-2sqrt(6))