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

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

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

vertices

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (4, 0), b = 4, a = 2

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

Refine

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

\frac{x ^{2}}{94} + \frac{y ^{2}}{36} = 1

x^{2} + 2 y^{2} - 2 x + 8 y + 5 = 0

\frac{x ^{2}}{49} + \frac{y ^{2}}{36} = 1

\frac{x ^{2}}{81} + \frac{y ^{2}}{64} = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{16} = 1

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