What is the foci ((x-5)^2)/4-((y+3)^2)/(16)=1 ?

The foci ((x-5)^2)/4-((y+3)^2)/(16)=1 is (5+2sqrt(5),-3),(5-2sqrt(5),-3)

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foci ((x-5)^2)/4-((y+3)^2)/(16)=1

foci

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{16} = 1

(5 + 25, - 3), (5 - 25, - 3)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 5 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (5, - 3), a = 2, b = 4

(5 + c, - 3), (5 - c, - 3)

Compute

c :

c = 2^{2} + 4^{2} : 25

(5 + 25, - 3), (5 - 25, - 3)

\frac{( x - 0 ) ^{2}}{16} - \frac{( y - 0 ) ^{2}}{9} = 1

\frac{( x - 4 ) ^{2}}{49} - \frac{( y + 6 ) ^{2}}{9} = 1

\frac{( x ^{2} )}{36} - \frac{( y ^{2} )}{25} = 1

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

y^{2} - 4 x^{2} < 1

What is the foci ((x-5)^2)/4-((y+3)^2)/(16)=1 ?
The foci ((x-5)^2)/4-((y+3)^2)/(16)=1 is (5+2sqrt(5),-3),(5-2sqrt(5),-3)