What is the foci ((x-4)^2)/4+((y-5)^2)/(25)=1 ?

The foci ((x-4)^2)/4+((y-5)^2)/(25)=1 is (4,5+sqrt(21)),(4,5-sqrt(21))

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foci ((x-4)^2)/4+((y-5)^2)/(25)=1

foci

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

(4, 5 + 21), (4, 5 - 21)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (4, 5), b = 5, a = 2

(4, 5 + c), (4, 5 - c)

Compute

c :

c = 5^{2} - 2^{2} : 21

(4, 5 + 21), (4, 5 - 21)

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

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

x^{2} + 9 y^{2} + 4 x = 5

4 x^{2} - 40 x + 2 y^{2} - 20 y = - 142

9 x^{2} + 7 y^{2} = 63

What is the foci ((x-4)^2)/4+((y-5)^2)/(25)=1 ?
The foci ((x-4)^2)/4+((y-5)^2)/(25)=1 is (4,5+sqrt(21)),(4,5-sqrt(21))