What is the vertices 25x^2+4y^2+200x-8y+304=0 ?

The vertices 25x^2+4y^2+200x-8y+304=0 is (-4,6),(-4,-4)

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vertices 25x^2+4y^2+200x-8y+304=0

vertices

25 x^{2} + 4 y^{2} + 200 x - 8 y + 304 = 0

(- 4, 6), (- 4, - 4)

Solution steps

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

Calculate ellipse properties

25 x^{2} + 4 y^{2} + 200 x - 8 y + 304 = 0 :

Ellipse with center

(h, k) = (- 4, 1), b = 5, a = 2

(- 4, 1 + 5), (- 4, 1 - 5)

Refine

(- 4, 6), (- 4, - 4)

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

3 x^{2} + 7 y^{2} - 21 = 0

\frac{( x - 3 ) ^{2}}{25} + (y - 2)^{2} = 1

x^{2} + 4 y^{2} + 2 x - 12 y + 6 = 0

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1

What is the vertices 25x^2+4y^2+200x-8y+304=0 ?
The vertices 25x^2+4y^2+200x-8y+304=0 is (-4,6),(-4,-4)