What is the foci ((x-2)^2)/4+((y-3)^2)/1 =1 ?

The foci ((x-2)^2)/4+((y-3)^2)/1 =1 is (2+sqrt(3),3),(2-sqrt(3),3)

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foci ((x-2)^2)/4+((y-3)^2)/1 =1

foci

\frac{( x - 2 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{1} = 1

(2 + 3, 3), (2 - 3, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{1} = 1 :

Ellipse with center

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

(2 + c, 3), (2 - c, 3)

Compute

c :

c = 2^{2} - 1^{2} : 3

(2 + 3, 3), (2 - 3, 3)

x^{2} + 4 y^{2} + 10 x + 9 y - 10 = 0

y^{2} = 1 - 4 x^{2}

16 x^{2} + 25 y^{2} = 1600

5 x^{2} + 9 y^{2} - 20 x - 18 y - 16 = 0

\frac{( x - 3 ) ^{2}}{64} + \frac{( y + 5 ) ^{2}}{81} = 1

What is the foci ((x-2)^2)/4+((y-3)^2)/1 =1 ?
The foci ((x-2)^2)/4+((y-3)^2)/1 =1 is (2+sqrt(3),3),(2-sqrt(3),3)