What is the foci ((x-1)^2)/(45)+(y^2)/(54)=1 ?

The foci ((x-1)^2)/(45)+(y^2)/(54)=1 is (1,3),(1,-3)

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foci ((x-1)^2)/(45)+(y^2)/(54)=1

foci

\frac{( x - 1 ) ^{2}}{45} + \frac{y ^{2}}{54} = 1

(1, 3), (1, - 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{45} + \frac{y ^{2}}{54} = 1 :

Ellipse with center

(h, k) = (1, 0), b = 36, a = 35

(1, 0 + c), (1, 0 - c)

Compute

c :

c = (36)^{2} - (35)^{2} : 3

(1, 0 + 3), (1, 0 - 3)

Refine

(1, 3), (1, - 3)

36 x^{2} + 5 y^{2} - 90 y - 495 = 0

16 x^{2} - 32 x + 9 y^{2} = 128

(\frac{x ^{2}}{15}) + (\frac{y ^{2}}{16}) = 1

9 x^{2} - 18 x + 25 y^{2} - 250 y + 409 = 0

25 x^{2} + 9 y^{2} + 50 x - 108 y + 124 = 0

What is the foci ((x-1)^2)/(45)+(y^2)/(54)=1 ?
The foci ((x-1)^2)/(45)+(y^2)/(54)=1 is (1,3),(1,-3)