What is the vertices (x^2)/3+(y^2)/5 =1 ?

The vertices (x^2)/3+(y^2)/5 =1 is (0,sqrt(5)),(0,-sqrt(5))

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vertices (x^2)/3+(y^2)/5 =1

vertices

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1

(0, 5), (0, - 5)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 5, a = 3

(0, 0 + 5), (0, 0 - 5)

Refine

(0, 5), (0, - 5)

4 x^{2} + 9 y^{2} - 36 = 0

9 x^{2} + 49 y^{2} - 144 x + 490 y + 1360 = 0

4 x^{2} + 9 y^{2} - 24 x + 72 y + 144 = 0

16 x^{2} + 25 y^{2} = 400

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

What is the vertices (x^2)/3+(y^2)/5 =1 ?
The vertices (x^2)/3+(y^2)/5 =1 is (0,sqrt(5)),(0,-sqrt(5))