What is the vertices 100x^2+36y^2=225 ?

The vertices 100x^2+36y^2=225 is (0, 5/2),(0,-5/2)

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vertices 100x^2+36y^2=225

vertices

100 x^{2} + 36 y^{2} = 225

(0, \frac{5}{2}), (0, - \frac{5}{2})

Solution steps

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

Calculate ellipse properties

100 x^{2} + 36 y^{2} = 225 :

Ellipse with center

(h, k) = (0, 0), b = \frac{5}{2}, a = \frac{3}{2}

(0, 0 + \frac{5}{2}), (0, 0 - \frac{5}{2})

Refine

(0, \frac{5}{2}), (0, - \frac{5}{2})

x^{2} + y^{2} + 2 x = 3 y

\frac{x ^{2}}{49} + \frac{y ^{2}}{169} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{64} = 1

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

9 x^{2} + y^{2} = 84

What is the vertices 100x^2+36y^2=225 ?
The vertices 100x^2+36y^2=225 is (0, 5/2),(0,-5/2)