What is the foci ((x-3)^2)/5+((y-1)^2)/(15)=1 ?

The foci ((x-3)^2)/5+((y-1)^2)/(15)=1 is (3,1+sqrt(10)),(3,1-sqrt(10))

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foci ((x-3)^2)/5+((y-1)^2)/(15)=1

foci

\frac{( x - 3 ) ^{2}}{5} + \frac{( y - 1 ) ^{2}}{15} = 1

(3, 1 + 10), (3, 1 - 10)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{5} + \frac{( y - 1 ) ^{2}}{15} = 1 :

Ellipse with center

(h, k) = (3, 1), b = 15, a = 5

(3, 1 + c), (3, 1 - c)

Compute

c :

c = 15^{2} - 5^{2} : 10

(3, 1 + 10), (3, 1 - 10)

\frac{x ^{2}}{100} + \frac{y ^{2}}{64} = 1

\frac{x ^{2}}{4} + \frac{y ^{2}}{16} = 1

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

\frac{x ^{2}}{25} + \frac{y ^{2}}{4} = 1

4 x^{2} + y^{2} = 16

What is the foci ((x-3)^2)/5+((y-1)^2)/(15)=1 ?
The foci ((x-3)^2)/5+((y-1)^2)/(15)=1 is (3,1+sqrt(10)),(3,1-sqrt(10))