What is the vertices (x^2)/(100)+(y^2)/(52)=1 ?

The vertices (x^2)/(100)+(y^2)/(52)=1 is (10,0),(-10,0)

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vertices (x^2)/(100)+(y^2)/(52)=1

vertices

\frac{x ^{2}}{100} + \frac{y ^{2}}{52} = 1

(10, 0), (- 10, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{100} + \frac{y ^{2}}{52} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 10, b = 213

(0 + 10, 0), (0 - 10, 0)

Refine

(10, 0), (- 10, 0)

x^{2} + 9 y^{2} + 16 x + 18 y + 64 = 0

x^{2} + 3 y^{2} + 6 x - 18 y + 33 = 0

\frac{( x + 1 ) ^{2}}{64} + \frac{( y - 6 ) ^{2}}{16} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{25} = 1

12.92 x^{2} + 0.07 y^{2} - 36 = 0

What is the vertices (x^2)/(100)+(y^2)/(52)=1 ?
The vertices (x^2)/(100)+(y^2)/(52)=1 is (10,0),(-10,0)