What is the foci ((y-3)^2)/7-((x+6)^2)/(42)=1 ?

The foci ((y-3)^2)/7-((x+6)^2)/(42)=1 is (-6,10),(-6,-4)

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foci ((y-3)^2)/7-((x+6)^2)/(42)=1

foci

\frac{( y - 3 ) ^{2}}{7} - \frac{( x + 6 ) ^{2}}{42} = 1

(- 6, 10), (- 6, - 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 3 ) ^{2}}{7} - \frac{( x + 6 ) ^{2}}{42} = 1 :

Up-down Hyperbola with

(h, k) = (- 6, 3), a = 7, b = 42

(- 6, 3 + c), (- 6, 3 - c)

Compute

c :

c = 7^{2} + 42^{2} : 7

(- 6, 3 + 7), (- 6, 3 - 7)

Refine

(- 6, 10), (- 6, - 4)

\frac{x ^{2}}{3} - \frac{y ^{2}}{3} = 1

2 x^{2} - 3 y^{2} = 6

49 y^{2} - x^{2} = 49

\frac{x ^{2}}{81} - \frac{y ^{2}}{1 6 ^{2}} = 1

9 x^{2} - 4 y^{2} + 72 x + 180 = 0

What is the foci ((y-3)^2)/7-((x+6)^2)/(42)=1 ?
The foci ((y-3)^2)/7-((x+6)^2)/(42)=1 is (-6,10),(-6,-4)