What is the foci (x/8)^2+(y/6)^2=1 ?

The foci (x/8)^2+(y/6)^2=1 is (2sqrt(7),0),(-2sqrt(7),0)

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foci (x/8)^2+(y/6)^2=1

foci

(\frac{x}{8})^{2} + (\frac{y}{6})^{2} = 1

(27, 0), (- 27, 0)

Solution steps

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

Calculate ellipse properties

(\frac{x}{8})^{2} + (\frac{y}{6})^{2} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 8, b = 6

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

Compute

c :

c = 8^{2} - 6^{2} : 27

(0 + 27, 0), (0 - 27, 0)

Refine

(27, 0), (- 27, 0)

\frac{x ^{2}}{484} + \frac{y ^{2}}{289} = 1

x^{2} + 9 y^{2} + 2 x - 54 y + 46 = 0

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

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

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

What is the foci (x/8)^2+(y/6)^2=1 ?
The foci (x/8)^2+(y/6)^2=1 is (2sqrt(7),0),(-2sqrt(7),0)