What is the foci 9x^2-4y^2+54x+16y+29=0 ?

The foci 9x^2-4y^2+54x+16y+29=0 is (-3+sqrt(13),2),(-3-sqrt(13),2)

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foci 9x^2-4y^2+54x+16y+29=0

foci

9 x^{2} - 4 y^{2} + 54 x + 16 y + 29 = 0

(- 3 + 13, 2), (- 3 - 13, 2)

Solution steps

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

Calculate Hyperbola properties

9 x^{2} - 4 y^{2} + 54 x + 16 y + 29 = 0 :

Right-left Hyperbola with

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

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

Compute

c :

c = 2^{2} + 3^{2} : 13

(- 3 + 13, 2), (- 3 - 13, 2)

8 y^{2} - 4 x^{2} = 4

\frac{x ^{2}}{49} - \frac{y ^{2}}{4} = 1

\frac{( y + \frac{7}{2} ) ^{2}}{25} - \frac{( x - \frac{9}{2} ) ^{2}}{25} = 1

y^{2} - 4 x^{2} = 16

\frac{y ^{2}}{4 ^{2}} - \frac{x ^{2}}{3 ^{2}} = 1

What is the foci 9x^2-4y^2+54x+16y+29=0 ?
The foci 9x^2-4y^2+54x+16y+29=0 is (-3+sqrt(13),2),(-3-sqrt(13),2)