What is the foci ((x-1)^2}{16}+\frac{(y-3)^2)/9 =1 ?

The foci ((x-1)^2}{16}+\frac{(y-3)^2)/9 =1 is (1+sqrt(7),3),(1-sqrt(7),3)

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foci ((x-1)^2}{16}+\frac{(y-3)^2)/9 =1

foci

\frac{( x - 1 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{9} = 1

(1 + 7, 3), (1 - 7, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (1, 3), a = 4, b = 3

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

Compute

c :

c = 4^{2} - 3^{2} : 7

(1 + 7, 3), (1 - 7, 3)

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 3 ) ^{2}}{9} = 1

\frac{( x + 3 ) ^{2}}{25} + \frac{( y - 5 ) ^{2}}{169} = 1

\frac{( x - 5 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{25} = 1

\frac{x ^{2}}{0.38 3 ^{2}} + \frac{y ^{2}}{0.374 8 ^{2}} = 1

\frac{x ^{2}}{140} + \frac{( y - 6 ) ^{2}}{4} = 1

What is the foci ((x-1)^2}{16}+\frac{(y-3)^2)/9 =1 ?
The foci ((x-1)^2}{16}+\frac{(y-3)^2)/9 =1 is (1+sqrt(7),3),(1-sqrt(7),3)