What is the vertices 25x^2+4y^2-50x+40y+25=0 ?

The vertices 25x^2+4y^2-50x+40y+25=0 is (1,0),(1,-10)

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vertices 25x^2+4y^2-50x+40y+25=0

vertices

25 x^{2} + 4 y^{2} - 50 x + 40 y + 25 = 0

(1, 0), (1, - 10)

Solution steps

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

Calculate ellipse properties

25 x^{2} + 4 y^{2} - 50 x + 40 y + 25 = 0 :

Ellipse with center

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

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

Refine

(1, 0), (1, - 10)

9 x^{2} + y^{2} + 18 x + 4 y + 4 = 0

\frac{x ^{2}}{9} + \frac{y ^{2}}{36} = 1

16 x^{2} + 25 y^{2} = 1600

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

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

What is the vertices 25x^2+4y^2-50x+40y+25=0 ?
The vertices 25x^2+4y^2-50x+40y+25=0 is (1,0),(1,-10)