What is the vertices (x^2)/(36)+(y^2)/(25)=1 ?

The vertices (x^2)/(36)+(y^2)/(25)=1 is (6,0),(-6,0)

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vertices (x^2)/(36)+(y^2)/(25)=1

vertices

\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1

(6, 0), (- 6, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 6, b = 5

(0 + 6, 0), (0 - 6, 0)

Refine

(6, 0), (- 6, 0)

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

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

4 x^{2} + y^{2} = 4 x

4 x^{2} + 4 x + y^{2} = 0

\frac{( x ^{2} )}{7} + \frac{( y ^{2} )}{16} = 1

What is the vertices (x^2)/(36)+(y^2)/(25)=1 ?
The vertices (x^2)/(36)+(y^2)/(25)=1 is (6,0),(-6,0)