What is the foci ((y+2)^2}{16}-\frac{(x+3)^2)/9 =1 ?

The foci ((y+2)^2}{16}-\frac{(x+3)^2)/9 =1 is (-3,3),(-3,-7)

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

foci

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

(- 3, 3), (- 3, - 7)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (- 3, - 2), a = 4, b = 3

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

Compute

c :

c = 4^{2} + 3^{2} : 5

(- 3, - 2 + 5), (- 3, - 2 - 5)

Refine

(- 3, 3), (- 3, - 7)

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

12 y^{2} - 4 x^{2} + 16 x + 72 y + 44 = 0

\frac{y ^{2}}{45} - \frac{x ^{2}}{36} = 1

y^{2} - 4 x^{2} + 16 x - 2 y = 31

\frac{y ^{2}}{12.00 6 ^{2}} - \frac{x ^{2}}{2.00 1 ^{2}} = 1

What is the foci ((y+2)^2}{16}-\frac{(x+3)^2)/9 =1 ?
The foci ((y+2)^2}{16}-\frac{(x+3)^2)/9 =1 is (-3,3),(-3,-7)