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

The vertices ((x+4)^2)/4+((y+1)^2)/(16)=1 is (-4,3),(-4,-5)

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

vertices

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

(- 4, 3), (- 4, - 5)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

(- 4, - 1 + 4), (- 4, - 1 - 4)

Refine

(- 4, 3), (- 4, - 5)

5 x^{2} + 9 y^{2} - 20 x + 54 y + 56 = 0

\frac{x ^{2}}{5} + \frac{y ^{2}}{7} = 1

9 x^{2} + 36 y^{2} = 324

36 x^{2} + y^{2} = 36

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

What is the vertices ((x+4)^2)/4+((y+1)^2)/(16)=1 ?
The vertices ((x+4)^2)/4+((y+1)^2)/(16)=1 is (-4,3),(-4,-5)