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

The vertices 9x^2+16y^2+96y=0 is (4,-3),(-4,-3)

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

vertices

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

(4, - 3), (- 4, - 3)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, - 3), a = 4, b = 3

(0 + 4, - 3), (0 - 4, - 3)

Refine

(4, - 3), (- 4, - 3)

49 x^{2} + 25 y^{2} - 1225 = 0

16 x^{2} + 64 y^{2} = 128

\frac{x ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

16 y^{2} + 25 x^{2} - 80 x + 36 y + 80 = 0

4 x^{2} + 9 y^{2} - 48 x + 72 y = - 144

What is the vertices 9x^2+16y^2+96y=0 ?
The vertices 9x^2+16y^2+96y=0 is (4,-3),(-4,-3)