What is the foci ((x-3)^2)/(16)-((y+2)^2)/(16)=1 ?

The foci ((x-3)^2)/(16)-((y+2)^2)/(16)=1 is (3+4sqrt(2),-2),(3-4sqrt(2),-2)

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foci ((x-3)^2)/(16)-((y+2)^2)/(16)=1

foci

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

(3 + 42, - 2), (3 - 42, - 2)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Compute

c :

c = 4^{2} + 4^{2} : 42

(3 + 42, - 2), (3 - 42, - 2)

x^{2} - 9 y^{2} - 4 x + 36 y - 41 = 0

4 y^{2} - 9 x^{2} + 40 y - 540 x - 8036 = 0

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

4 x^{2} - 4 y^{2} + 20 x - 16 y + 25 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{81} = 1

What is the foci ((x-3)^2)/(16)-((y+2)^2)/(16)=1 ?
The foci ((x-3)^2)/(16)-((y+2)^2)/(16)=1 is (3+4sqrt(2),-2),(3-4sqrt(2),-2)