What is the foci 9x^2+4y^2-54x+40y+145=0 ?

The foci 9x^2+4y^2-54x+40y+145=0 is (3,-5+sqrt(5)),(3,-5-sqrt(5))

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foci 9x^2+4y^2-54x+40y+145=0

foci

9 x^{2} + 4 y^{2} - 54 x + 40 y + 145 = 0

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

Solution steps

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

Calculate ellipse properties

9 x^{2} + 4 y^{2} - 54 x + 40 y + 145 = 0 :

Ellipse with center

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

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

Compute

c :

c = 3^{2} - 2^{2} : 5

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

9 x^{2} + y^{2} + 36 x - 2 y + 1 = 0

\frac{( x ) ^{2}}{46} + \frac{( y + 8 ) ^{2}}{26} = 1

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 5 ) ^{2}}{25} = 1

81 x^{2} + 36 y^{2} = 2916

5 x^{2} + 7 y^{2} = 70

What is the foci 9x^2+4y^2-54x+40y+145=0 ?
The foci 9x^2+4y^2-54x+40y+145=0 is (3,-5+sqrt(5)),(3,-5-sqrt(5))