What is the foci ((x-1)^2)/(25)+(y^2)/(16)=1 ?

The foci ((x-1)^2)/(25)+(y^2)/(16)=1 is (4,0),(-2,0)

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foci ((x-1)^2)/(25)+(y^2)/(16)=1

foci

\frac{( x - 1 ) ^{2}}{25} + \frac{y ^{2}}{16} = 1

(4, 0), (- 2, 0)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{25} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (1, 0), a = 5, b = 4

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

Compute

c :

c = 5^{2} - 4^{2} : 3

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

Refine

(4, 0), (- 2, 0)

\frac{x ^{2}}{576} + \frac{y ^{2}}{676} = 1

x^{2} + 3 y^{2} + 6 x + 6 = 0

16 x^{2} + 25 y^{2} - 48 x + 100 y - 264 = 0

9 x^{2} + 25 y^{2} + 54 x + 100 y - 44 = 0

15 x^{2} + 16 y^{2} = 240

What is the foci ((x-1)^2)/(25)+(y^2)/(16)=1 ?
The foci ((x-1)^2)/(25)+(y^2)/(16)=1 is (4,0),(-2,0)