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

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

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

foci

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

(42, 4), (- 42, 4)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, 4), a = 4, b = 4

(0 + c, 4), (0 - c, 4)

Compute

c :

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

(0 + 42, 4), (0 - 42, 4)

Refine

(42, 4), (- 42, 4)

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

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

\frac{( x + 2 ) ^{2}}{9} - \frac{( y - 1 ) ^{2}}{169} = 1

y^{2} - 49 x^{2} - 14 y - 490 x - 1125 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{4} = 1

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