What is the vertices (x^2)/5+(y^2)/7 =1 ?

The vertices (x^2)/5+(y^2)/7 =1 is (0,sqrt(7)),(0,-sqrt(7))

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vertices (x^2)/5+(y^2)/7 =1

vertices

\frac{x ^{2}}{5} + \frac{y ^{2}}{7} = 1

(0, 7), (0, - 7)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{5} + \frac{y ^{2}}{7} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 7, a = 5

(0, 0 + 7), (0, 0 - 7)

Refine

(0, 7), (0, - 7)

9 x^{2} + 36 y^{2} = 324

36 x^{2} + y^{2} = 36

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

\frac{x ^{2}}{2 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

What is the vertices (x^2)/5+(y^2)/7 =1 ?
The vertices (x^2)/5+(y^2)/7 =1 is (0,sqrt(7)),(0,-sqrt(7))