What is the foci ((x-2)^2)/9+((y+1)^2)/(36)=1 ?

The foci ((x-2)^2)/9+((y+1)^2)/(36)=1 is (2,-1+3sqrt(3)),(2,-1-3sqrt(3))

Popular >

foci ((x-2)^2)/9+((y+1)^2)/(36)=1

foci

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{36} = 1

(2, - 1 + 33), (2, - 1 - 33)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{9} + \frac{( y + 1 ) ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (2, - 1), b = 6, a = 3

(2, - 1 + c), (2, - 1 - c)

Compute

c :

c = 6^{2} - 3^{2} : 33

(2, - 1 + 33), (2, - 1 - 33)

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 2 ) ^{2}}{4} = 1

\frac{x ^{2}}{63500} + \frac{y ^{2}}{50900} = 1

\frac{( x - 6 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{25} = 1

4 x^{2} + 25 y^{2} + 24 x + 250 y = - 561

4 x^{2} + 36 y^{2} + 24 x + 432 y = - 1188

What is the foci ((x-2)^2)/9+((y+1)^2)/(36)=1 ?
The foci ((x-2)^2)/9+((y+1)^2)/(36)=1 is (2,-1+3sqrt(3)),(2,-1-3sqrt(3))