What is the vertices (x^2)/(144)+(y^2)/(225)=1 ?

The vertices (x^2)/(144)+(y^2)/(225)=1 is (0,15),(0,-15)

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vertices (x^2)/(144)+(y^2)/(225)=1

vertices

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1

(0, 15), (0, - 15)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 15, a = 12

(0, 0 + 15), (0, 0 - 15)

Refine

(0, 15), (0, - 15)

\frac{( x + 6 ) ^{2}}{7} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{5 x ^{2} + 8 y ^{2} - 1}{11 x ^{2} - 13 y ^{2}} = 0

\frac{y ^{2}}{16} + \frac{x ^{2}}{49} = 1

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

4 x^{2} + 81 y^{2} = 324

What is the vertices (x^2)/(144)+(y^2)/(225)=1 ?
The vertices (x^2)/(144)+(y^2)/(225)=1 is (0,15),(0,-15)