What is the vertices (x^2)/(56)+(y^2)/(36)=1 ?

The vertices (x^2)/(56)+(y^2)/(36)=1 is (2sqrt(14),0),(-2sqrt(14),0)

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vertices (x^2)/(56)+(y^2)/(36)=1

vertices

\frac{x ^{2}}{56} + \frac{y ^{2}}{36} = 1

(214, 0), (- 214, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{56} + \frac{y ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 214, b = 6

(0 + 214, 0), (0 - 214, 0)

Refine

(214, 0), (- 214, 0)

16 (x - 1)^{2} + 25 (y - 4)^{2} - 400 = 0

9 x^{2} + 4 y^{2} + 36 x + 24 y + 36 = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{169} = 1

6 x^{2} + 9 y^{2} - 18 y - 45 = 0

7 x^{2} + 16 y^{2} - 112 = 0

What is the vertices (x^2)/(56)+(y^2)/(36)=1 ?
The vertices (x^2)/(56)+(y^2)/(36)=1 is (2sqrt(14),0),(-2sqrt(14),0)