What is the foci ((x-3)^2)/(100)+((y-1)^2)/(64)=1 ?

The foci ((x-3)^2)/(100)+((y-1)^2)/(64)=1 is (9,1),(-3,1)

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foci ((x-3)^2)/(100)+((y-1)^2)/(64)=1

foci

\frac{( x - 3 ) ^{2}}{100} + \frac{( y - 1 ) ^{2}}{64} = 1

(9, 1), (- 3, 1)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{100} + \frac{( y - 1 ) ^{2}}{64} = 1 :

Ellipse with center

(h, k) = (3, 1), a = 10, b = 8

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

Compute

c :

c = 1 0^{2} - 8^{2} : 6

(3 + 6, 1), (3 - 6, 1)

Refine

(9, 1), (- 3, 1)

25 x^{2} + y^{2} - 4 y - 21 = 0

16 x^{2} + 25 y^{2} - 64 x + 150 y = - 279

\frac{( x - 12 ) ^{2}}{100} + \frac{( y + 7 ) ^{2}}{36} = 1

\frac{x ^{2}}{0.38 4 ^{2}} + \frac{y ^{2}}{0.375 8 ^{2}} = 1

4 x^{2} + y^{2} + 6 y + 5 = 0

What is the foci ((x-3)^2)/(100)+((y-1)^2)/(64)=1 ?
The foci ((x-3)^2)/(100)+((y-1)^2)/(64)=1 is (9,1),(-3,1)