What is the vertices 16x^2+25y^2-64x+150y-111=0 ?

The vertices 16x^2+25y^2-64x+150y-111=0 is (7,-3),(-3,-3)

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vertices 16x^2+25y^2-64x+150y-111=0

vertices

16 x^{2} + 25 y^{2} - 64 x + 150 y - 111 = 0

(7, - 3), (- 3, - 3)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} - 64 x + 150 y - 111 = 0 :

Ellipse with center

(h, k) = (2, - 3), a = 5, b = 4

(2 + 5, - 3), (2 - 5, - 3)

Refine

(7, - 3), (- 3, - 3)

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

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{49} = 1

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

What is the vertices 16x^2+25y^2-64x+150y-111=0 ?
The vertices 16x^2+25y^2-64x+150y-111=0 is (7,-3),(-3,-3)