What is the foci x^2+16y^2=1 ?

The foci x^2+16y^2=1 is ((sqrt(15))/4 ,0),(-(sqrt(15))/4 ,0)

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foci x^2+16y^2=1

foci

x^{2} + 16 y^{2} = 1

(\frac{15}{4}, 0), (- \frac{15}{4}, 0)

Solution steps

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

Calculate ellipse properties

x^{2} + 16 y^{2} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 1, b = \frac{1}{4}

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

Compute

c :

c = 1^{2} - (\frac{1}{4})^{2} : \frac{15}{4}

(0 + \frac{15}{4}, 0), (0 - \frac{15}{4}, 0)

Refine

(\frac{15}{4}, 0), (- \frac{15}{4}, 0)

28 x^{2} + 64 y^{2} - 56 x - 420 = 0

\frac{( x + 1 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{10} = 1

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

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

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

What is the foci x^2+16y^2=1 ?
The foci x^2+16y^2=1 is ((sqrt(15))/4 ,0),(-(sqrt(15))/4 ,0)