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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

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

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

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

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