What is the vertices 2x^2+y^2-98=0 ?

The vertices 2x^2+y^2-98=0 is (0,7sqrt(2)),(0,-7sqrt(2))

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vertices 2x^2+y^2-98=0

vertices

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

(0, 72), (0, - 72)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), b = 72, a = 7

(0, 0 + 72), (0, 0 - 72)

Refine

(0, 72), (0, - 72)

3 x^{2} - 6 x + y^{2} + 4 y - 11 = 0

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

\frac{x ^{2}}{144} + \frac{y ^{2}}{169} = 1

9 x^{2} + y^{2} - 18 x - 6 y + 9 = 0

\frac{( x - 0 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{49} = 1

What is the vertices 2x^2+y^2-98=0 ?
The vertices 2x^2+y^2-98=0 is (0,7sqrt(2)),(0,-7sqrt(2))