What is the foci ((x-4)^2)/(5^2)+((y-1)^2)/(2^2)=1 ?

The foci ((x-4)^2)/(5^2)+((y-1)^2)/(2^2)=1 is (4+sqrt(21),1),(4-sqrt(21),1)

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foci ((x-4)^2)/(5^2)+((y-1)^2)/(2^2)=1

foci

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

(4 + 21, 1), (4 - 21, 1)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (4, 1), a = 5, b = 2

(4 + c, 1), (4 - c, 1)

Compute

c :

c = 5^{2} - 2^{2} : 21

(4 + 21, 1), (4 - 21, 1)

16 x^{2} + 25 y^{2} + 100 y = 300

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{9} = 1

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

(\frac{x}{2.5})^{2} + (\frac{y}{3})^{2} = 1

9 x^{2} + 4 y^{2} + 54 x - 8 y + 49 = 0

What is the foci ((x-4)^2)/(5^2)+((y-1)^2)/(2^2)=1 ?
The foci ((x-4)^2)/(5^2)+((y-1)^2)/(2^2)=1 is (4+sqrt(21),1),(4-sqrt(21),1)