What is the vertices 4x^2+9y^2=25 ?

The vertices 4x^2+9y^2=25 is (5/2 ,0),(-5/2 ,0)

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vertices 4x^2+9y^2=25

vertices

4 x^{2} + 9 y^{2} = 25

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

Solution steps

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

Calculate ellipse properties

4 x^{2} + 9 y^{2} = 25 :

Ellipse with center

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

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

Refine

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

\frac{( x + 3 ) ^{2}}{26} + \frac{( y - 4 ) ^{2}}{16} = 1

\frac{x ^{2}}{8} + \frac{y ^{2}}{4} = 1

4 x^{2} + y^{2} - 32 x + 16 y + 124 = 0

(x - 2)^{2} + 4 (y - 3)^{2} = 4

x^{2} + 4 y^{2} - 6 x + 20 y - 2 = 0

What is the vertices 4x^2+9y^2=25 ?
The vertices 4x^2+9y^2=25 is (5/2 ,0),(-5/2 ,0)