What is the foci 25x^2-16y^2+250x+32y+109=0 ?

The foci 25x^2-16y^2+250x+32y+109=0 is ((-10+sqrt(205))/2 ,1),((-10-sqrt(205))/2 ,1)

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foci 25x^2-16y^2+250x+32y+109=0

foci

25 x^{2} - 16 y^{2} + 250 x + 32 y + 109 = 0

(\frac{- 10 + 205}{2}, 1), (\frac{- 10 - 205}{2}, 1)

Solution steps

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

Calculate Hyperbola properties

25 x^{2} - 16 y^{2} + 250 x + 32 y + 109 = 0 :

Right-left Hyperbola with

(h, k) = (- 5, 1), a = 25, b = \frac{5 5}{2}

(- 5 + c, 1), (- 5 - c, 1)

Compute

c :

c = (25)^{2} + (\frac{5 5}{2})^{2} : \frac{205}{2}

(- 5 + \frac{205}{2}, 1), (- 5 - \frac{205}{2}, 1)

Refine

(\frac{- 10 + 205}{2}, 1), (\frac{- 10 - 205}{2}, 1)

\frac{( x + 2 ) ^{2}}{25} - \frac{( y - 1 ) ^{2}}{169} = 1

\frac{( x - 2 ) ^{2}}{5} - \frac{( y + 1 ) ^{2}}{9} = 1

\frac{y ^{2}}{4} - \frac{x ^{2}}{1} = 1

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

5 x^{2} - 4 y^{2} + 80 = 0

What is the foci 25x^2-16y^2+250x+32y+109=0 ?
The foci 25x^2-16y^2+250x+32y+109=0 is ((-10+sqrt(205))/2 ,1),((-10-sqrt(205))/2 ,1)