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

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

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

vertices

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

(- 1, - 4), (- 3, - 4)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 2, - 4), a = 1, b = \frac{1}{2}

(- 2 + 1, - 4), (- 2 - 1, - 4)

Refine

(- 1, - 4), (- 3, - 4)

\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{7} = 1

x = - 5 y^{2}

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0

y^{2} = - 16 x

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