What is the foci ((x-9)^2)/4+((y+10)^2)/(81)=1 ?

The foci ((x-9)^2)/4+((y+10)^2)/(81)=1 is (9,-10+sqrt(77)),(9,-10-sqrt(77))

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foci ((x-9)^2)/4+((y+10)^2)/(81)=1

foci

\frac{( x - 9 ) ^{2}}{4} + \frac{( y + 10 ) ^{2}}{81} = 1

(9, - 10 + 77), (9, - 10 - 77)

Solution steps

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

Calculate ellipse properties

\frac{( x - 9 ) ^{2}}{4} + \frac{( y + 10 ) ^{2}}{81} = 1 :

Ellipse with center

(h, k) = (9, - 10), b = 9, a = 2

(9, - 10 + c), (9, - 10 - c)

Compute

c :

c = 9^{2} - 2^{2} : 77

(9, - 10 + 77), (9, - 10 - 77)

18 x^{2} + 8 y^{2} + 36 x - 32 y - 22 = 0

\frac{x ^{2}}{32} + \frac{y ^{2}}{9} = 1

4 x^{2} + 25 y^{2} - 24 x + 250 y + 561 = 0

14 x^{2} - 70 = - 5 y^{2}

- \frac{( y - 1 ) ^{2}}{144} - \frac{( x + 2 ) ^{2}}{9} = 1

What is the foci ((x-9)^2)/4+((y+10)^2)/(81)=1 ?
The foci ((x-9)^2)/4+((y+10)^2)/(81)=1 is (9,-10+sqrt(77)),(9,-10-sqrt(77))