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

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

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vertices (x^2)/(16)+(y-4)^2=1

vertices

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

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + (y - 4)^{2} = 1 :

Ellipse with center

(h, k) = (0, 4), a = 4, b = 1

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

Refine

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

3 x^{2} + y^{2} - 24 x + 39 = 0

25 x^{2} + 16 y^{2} = 400

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} > 1

12 x^{2} + 20 y^{2} - 12 x + 40 y - 37 = 0

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 7 ) ^{2}}{16} = 1

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