What is the vertices 25(x+2)^2+4(y-3)^2=100 ?

The vertices 25(x+2)^2+4(y-3)^2=100 is (-2,8),(-2,-2)

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vertices 25(x+2)^2+4(y-3)^2=100

vertices

25 (x + 2)^{2} + 4 (y - 3)^{2} = 100

(- 2, 8), (- 2, - 2)

Solution steps

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

Calculate ellipse properties

25 (x + 2)^{2} + 4 (y - 3)^{2} = 100 :

Ellipse with center

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

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

Refine

(- 2, 8), (- 2, - 2)

4 x^{2} + y^{2} + 6 y + 5 = 0

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{49} = 1

(\frac{x}{7})^{2} + (\frac{y}{5})^{2} = 1

36 x^{2} + 25 y^{2} - 50 y - 875 = 0

4 (x - 5)^{2} + 3 (y - 5)^{2} = 192

What is the vertices 25(x+2)^2+4(y-3)^2=100 ?
The vertices 25(x+2)^2+4(y-3)^2=100 is (-2,8),(-2,-2)