What is the vertices (x^2)/7+(y^2)/(16)=1 ?

The vertices (x^2)/7+(y^2)/(16)=1 is (0,4),(0,-4)

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vertices (x^2)/7+(y^2)/(16)=1

vertices

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{7} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 4, a = 7

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

Refine

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

\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1

x^{2} - y^{2} = 14

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

16 x^{2} + 192 x + 576 + 25 y^{2} = 1600

x^{2} + y^{2} - 10 x - 10 y + 25 = 0

What is the vertices (x^2)/7+(y^2)/(16)=1 ?
The vertices (x^2)/7+(y^2)/(16)=1 is (0,4),(0,-4)