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

The vertices ((x+2)^2)/4+((y-4)^2)/(16)=1 is (-2,8),(-2,0)

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

vertices

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

(- 2, 8), (- 2, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Refine

(- 2, 8), (- 2, 0)

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1

4 x^{2} + 9 y^{2} - 36 = 0

9 x^{2} + 49 y^{2} - 144 x + 490 y + 1360 = 0

4 x^{2} + 9 y^{2} - 24 x + 72 y + 144 = 0

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

What is the vertices ((x+2)^2)/4+((y-4)^2)/(16)=1 ?
The vertices ((x+2)^2)/4+((y-4)^2)/(16)=1 is (-2,8),(-2,0)