What is the foci (x^2)/(144)+(y^2)/(225)=1 ?

The foci (x^2)/(144)+(y^2)/(225)=1 is (0,9),(0,-9)

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foci (x^2)/(144)+(y^2)/(225)=1

foci

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1

(0, 9), (0, - 9)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 15, a = 12

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

Compute

c :

c = 1 5^{2} - 1 2^{2} : 9

(0, 0 + 9), (0, 0 - 9)

Refine

(0, 9), (0, - 9)

4 x^{2} + 25 y^{2} + 40 x - 150 y + 225 = 0

\frac{x ^{2}}{\frac{81}{4}} + \frac{y ^{2}}{\frac{25}{16}} = 1

2 x^{2} + 3 y^{2} + 4 x - 1 = 0

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

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

What is the foci (x^2)/(144)+(y^2)/(225)=1 ?
The foci (x^2)/(144)+(y^2)/(225)=1 is (0,9),(0,-9)