What is the foci 16(x-1)^2+9(y+2)^2=144 ?

The foci 16(x-1)^2+9(y+2)^2=144 is (1,-2+sqrt(7)),(1,-2-sqrt(7))

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foci 16(x-1)^2+9(y+2)^2=144

foci

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

(1, - 2 + 7), (1, - 2 - 7)

Solution steps

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

Calculate ellipse properties

16 (x - 1)^{2} + 9 (y + 2)^{2} = 144 :

Ellipse with center

(h, k) = (1, - 2), b = 4, a = 3

(1, - 2 + c), (1, - 2 - c)

Compute

c :

c = 4^{2} - 3^{2} : 7

(1, - 2 + 7), (1, - 2 - 7)

\frac{( x - 5 ) ^{2}}{3} + \frac{( y - 7 ) ^{2}}{6} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{50} = 1

\frac{x ^{2}}{49} + \frac{y ^{2}}{81} = 1

25 x^{2} + 16 y^{2} + 100 x + 96 y - 156 = 0

9 x^{2} + 16 y^{2} + 64 y - 80 = 0

What is the foci 16(x-1)^2+9(y+2)^2=144 ?
The foci 16(x-1)^2+9(y+2)^2=144 is (1,-2+sqrt(7)),(1,-2-sqrt(7))