What is the foci (x^2)/(16)+((y-1)^2)/(49)=1 ?

The foci (x^2)/(16)+((y-1)^2)/(49)=1 is (0,1+sqrt(33)),(0,1-sqrt(33))

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foci (x^2)/(16)+((y-1)^2)/(49)=1

foci

\frac{x ^{2}}{16} + \frac{( y - 1 ) ^{2}}{49} = 1

(0, 1 + 33), (0, 1 - 33)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{( y - 1 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, 1), b = 7, a = 4

(0, 1 + c), (0, 1 - c)

Compute

c :

c = 7^{2} - 4^{2} : 33

(0, 1 + 33), (0, 1 - 33)

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{4} = 1

4 x^{2} + 9 y^{2} + 12 x - 108 y + 297 = 0

9 x^{(2)} + 16 y^{(2)} - 54 x + 64 y + 1 = 0

\frac{( x + 4 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

2 x^{2} + 4 y^{2} + 8 x + 9 y - 3 = 0

What is the foci (x^2)/(16)+((y-1)^2)/(49)=1 ?
The foci (x^2)/(16)+((y-1)^2)/(49)=1 is (0,1+sqrt(33)),(0,1-sqrt(33))