What is the vertices ((x-2)^2)/(16)+(y-5)^2=1 ?

The vertices ((x-2)^2)/(16)+(y-5)^2=1 is (6,5),(-2,5)

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vertices ((x-2)^2)/(16)+(y-5)^2=1

vertices

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

(6, 5), (- 2, 5)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, 5), a = 4, b = 1

(2 + 4, 5), (2 - 4, 5)

Refine

(6, 5), (- 2, 5)

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

\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

What is the vertices ((x-2)^2)/(16)+(y-5)^2=1 ?
The vertices ((x-2)^2)/(16)+(y-5)^2=1 is (6,5),(-2,5)