What is the vertices 25x^2+16y^2-50x+64y-311=0 ?

The vertices 25x^2+16y^2-50x+64y-311=0 is (1,3),(1,-7)

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vertices 25x^2+16y^2-50x+64y-311=0

vertices

25 x^{2} + 16 y^{2} - 50 x + 64 y - 311 = 0

(1, 3), (1, - 7)

Solution steps

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

Calculate ellipse properties

25 x^{2} + 16 y^{2} - 50 x + 64 y - 311 = 0 :

Ellipse with center

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

(1, - 2 + 5), (1, - 2 - 5)

Refine

(1, 3), (1, - 7)

16 x^{2} + 36 y^{2} - 256 x + 448 = 0

9 x^{2} + 25 y^{2} - 36 x - 50 y + 60 = 0

\frac{( x + 1 ) ^{2}}{49} + \frac{( y - 2 ) ^{2}}{48} = 1

3 x^{2} + 4 y^{2} = 48

2 x^{2} + y^{2} - 98 = 0

What is the vertices 25x^2+16y^2-50x+64y-311=0 ?
The vertices 25x^2+16y^2-50x+64y-311=0 is (1,3),(1,-7)