What is the vertices 4x^2+y^2+6y+5=0 ?

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

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vertices 4x^2+y^2+6y+5=0

vertices

4 x^{2} + y^{2} + 6 y + 5 = 0

(0, - 1), (0, - 5)

Solution steps

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

Calculate ellipse properties

4 x^{2} + y^{2} + 6 y + 5 = 0 :

Ellipse with center

(h, k) = (0, - 3), b = 2, a = 1

(0, - 3 + 2), (0, - 3 - 2)

Refine

(0, - 1), (0, - 5)

\frac{( x + 2 ) ^{2}}{16} + \frac{y ^{2}}{49} = 1

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

36 x^{2} + 25 y^{2} - 50 y - 875 = 0

4 (x - 5)^{2} + 3 (y - 5)^{2} = 192

9 x^{2} + 16 y^{2} + 36 x - 160 y + 292 = 0

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