What is the vertices 49x^2+25y^2-196x+250y-404=0 ?

The vertices 49x^2+25y^2-196x+250y-404=0 is (2,2),(2,-12)

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vertices 49x^2+25y^2-196x+250y-404=0

vertices

49 x^{2} + 25 y^{2} - 196 x + 250 y - 404 = 0

(2, 2), (2, - 12)

Solution steps

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

Calculate ellipse properties

49 x^{2} + 25 y^{2} - 196 x + 250 y - 404 = 0 :

Ellipse with center

(h, k) = (2, - 5), b = 7, a = 5

(2, - 5 + 7), (2, - 5 - 7)

Refine

(2, 2), (2, - 12)

\frac{( x - 3 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{25} = 1

\frac{x ^{2}}{64} + \frac{y ^{2}}{100} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

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

What is the vertices 49x^2+25y^2-196x+250y-404=0 ?
The vertices 49x^2+25y^2-196x+250y-404=0 is (2,2),(2,-12)