What is the vertices 4x^2+9y^2-36=0 ?

The vertices 4x^2+9y^2-36=0 is (3,0),(-3,0)

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vertices 4x^2+9y^2-36=0

vertices

4 x^{2} + 9 y^{2} - 36 = 0

(3, 0), (- 3, 0)

Solution steps

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

Calculate ellipse properties

4 x^{2} + 9 y^{2} - 36 = 0 :

Ellipse with center

(h, k) = (0, 0), a = 3, b = 2

(0 + 3, 0), (0 - 3, 0)

Refine

(3, 0), (- 3, 0)

9 x^{2} + 49 y^{2} - 144 x + 490 y + 1360 = 0

4 x^{2} + 9 y^{2} - 24 x + 72 y + 144 = 0

16 x^{2} + 25 y^{2} = 400

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

9 x^{2} + 4 y^{2} + 18 x - 8 y - 131 = 0

What is the vertices 4x^2+9y^2-36=0 ?
The vertices 4x^2+9y^2-36=0 is (3,0),(-3,0)