What is the vertices (x^2)/8+(y^2)/5 =1 ?

The vertices (x^2)/8+(y^2)/5 =1 is (2sqrt(2),0),(-2sqrt(2),0)

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vertices (x^2)/8+(y^2)/5 =1

vertices

\frac{x ^{2}}{8} + \frac{y ^{2}}{5} = 1

(22, 0), (- 22, 0)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

\frac{x ^{2}}{8} + \frac{y ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 22, b = 5

(0 + 22, 0), (0 - 22, 0)

Refine

(22, 0), (- 22, 0)

9 x^{2} + 16 y^{2} - 144 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1

16 x^{2} + 9 y^{2} - 64 x - 54 y + 1 = 0

\frac{( x - 4 ) ^{2}}{4} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{94} + \frac{y ^{2}}{36} = 1

What is the vertices (x^2)/8+(y^2)/5 =1 ?
The vertices (x^2)/8+(y^2)/5 =1 is (2sqrt(2),0),(-2sqrt(2),0)