What is the foci ((x-3)^2)/9-((y+10)^2)/(100)=1 ?

The foci ((x-3)^2)/9-((y+10)^2)/(100)=1 is (3+sqrt(109),-10),(3-sqrt(109),-10)

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foci ((x-3)^2)/9-((y+10)^2)/(100)=1

foci

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 10 ) ^{2}}{100} = 1

(3 + 109, - 10), (3 - 109, - 10)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 10 ) ^{2}}{100} = 1 :

Right-left Hyperbola with

(h, k) = (3, - 10), a = 3, b = 10

(3 + c, - 10), (3 - c, - 10)

Compute

c :

c = 3^{2} + 1 0^{2} : 109

(3 + 109, - 10), (3 - 109, - 10)

4 x^{2} - 12 y^{2} - 24 x + 24 y + 51 = 0

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

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

What is the foci ((x-3)^2)/9-((y+10)^2)/(100)=1 ?
The foci ((x-3)^2)/9-((y+10)^2)/(100)=1 is (3+sqrt(109),-10),(3-sqrt(109),-10)