What is the vertices 12.92x^2+0.07y^2-36=0 ?

The vertices 12.92x^2+0.07y^2-36=0 is (0,22.67786…),(0,-22.67786…)

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vertices 12.92x^2+0.07y^2-36=0

vertices

12.92 x^{2} + 0.07 y^{2} - 36 = 0

(0, 22.67786 \dots), (0, - 22.67786 \dots)

Solution steps

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

Calculate ellipse properties

12.92 x^{2} + 0.07 y^{2} - 36 = 0 :

Ellipse with center

(h, k) = (0, 0), b = 22.67786 \dots, a = 1.66924 \dots

(0, 0 + 22.67786 \dots), (0, 0 - 22.67786 \dots)

Refine

(0, 22.67786 \dots), (0, - 22.67786 \dots)

15 x^{2} + 16 y^{2} = 240

\frac{x ^{2}}{49} + \frac{( y - 3 ) ^{2}}{16} = 1

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400

3 x^{2} + 4 y^{2} - 12 x = 36

\frac{x ^{2}}{64} + \frac{y ^{2}}{4} = 1

What is the vertices 12.92x^2+0.07y^2-36=0 ?
The vertices 12.92x^2+0.07y^2-36=0 is (0,22.67786…),(0,-22.67786…)