What is the foci x^2+5y^2-2x+10y=-1 ?

The foci x^2+5y^2-2x+10y=-1 is (3,-1),(-1,-1)

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foci x^2+5y^2-2x+10y=-1

foci

x^{2} + 5 y^{2} - 2 x + 10 y = - 1

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

Solution steps

(h + c, k), (h - c, k)

Calculate ellipse properties

x^{2} + 5 y^{2} - 2 x + 10 y = - 1 :

Ellipse with center

(h, k) = (1, - 1), a = 5, b = 1

(1 + c, - 1), (1 - c, - 1)

Compute

c :

c = 5^{2} - 1^{2} : 2

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

Refine

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

(\frac{x}{8})^{2} + (\frac{y}{6})^{2} = 1

\frac{x ^{2}}{484} + \frac{y ^{2}}{289} = 1

x^{2} + 9 y^{2} + 2 x - 54 y + 46 = 0

\frac{( x - 1 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{9} = 1

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{9} = 1

What is the foci x^2+5y^2-2x+10y=-1 ?
The foci x^2+5y^2-2x+10y=-1 is (3,-1),(-1,-1)