What is the vertices (x^2)/(16)+(y^2)/(12)=1 ?

The vertices (x^2)/(16)+(y^2)/(12)=1 is (4,0),(-4,0)

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vertices (x^2)/(16)+(y^2)/(12)=1

vertices

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1

(4, 0), (- 4, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{y ^{2}}{12} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 4, b = 23

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

41 x^{2} + 16 y^{2} + 246 x - 192 y + 289 = 0

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

\frac{x ^{2}}{4} + \frac{y ^{2}}{25} = 1

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

9 x^{2} - 18 x + 4 y^{2} = 27

What is the vertices (x^2)/(16)+(y^2)/(12)=1 ?
The vertices (x^2)/(16)+(y^2)/(12)=1 is (4,0),(-4,0)