What is the vertices x^2+4y^2+6x+16y+21=0 ?

The vertices x^2+4y^2+6x+16y+21=0 is (-1,-2),(-5,-2)

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vertices x^2+4y^2+6x+16y+21=0

vertices

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0

(- 1, - 2), (- 5, - 2)

Solution steps

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

Calculate ellipse properties

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0 :

Ellipse with center

(h, k) = (- 3, - 2), a = 2, b = 1

(- 3 + 2, - 2), (- 3 - 2, - 2)

Refine

(- 1, - 2), (- 5, - 2)

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

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1

4 x^{2} + 25 y^{2} - 100 = 0

16 x^{2} + 25 y^{2} - 400 = 0

\frac{x ^{2}}{9} + y^{2} = 1

What is the vertices x^2+4y^2+6x+16y+21=0 ?
The vertices x^2+4y^2+6x+16y+21=0 is (-1,-2),(-5,-2)