What is the vertices 3x^2+7y^2=63 ?

The vertices 3x^2+7y^2=63 is (sqrt(21),0),(-sqrt(21),0)

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vertices 3x^2+7y^2=63

vertices

3 x^{2} + 7 y^{2} = 63

(21, 0), (- 21, 0)

Solution steps

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

Calculate ellipse properties

3 x^{2} + 7 y^{2} = 63 :

Ellipse with center

(h, k) = (0, 0), a = 21, b = 3

(0 + 21, 0), (0 - 21, 0)

Refine

(21, 0), (- 21, 0)

\frac{x ^{2}}{4} + \frac{y ^{2}}{16} = 1

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

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

25 x^{2} + 4 y^{2} - 50 x + 40 y + 25 = 0

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

What is the vertices 3x^2+7y^2=63 ?
The vertices 3x^2+7y^2=63 is (sqrt(21),0),(-sqrt(21),0)