What is the foci 9x^2+16y^2=1 ?

The foci 9x^2+16y^2=1 is ((sqrt(7))/(12),0),(-(sqrt(7))/(12),0)

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foci 9x^2+16y^2=1

foci

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

(\frac{7}{12}, 0), (- \frac{7}{12}, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = \frac{1}{3}, b = \frac{1}{4}

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

Compute

c :

c = (\frac{1}{3})^{2} - (\frac{1}{4})^{2} : \frac{7}{12}

(0 + \frac{7}{12}, 0), (0 - \frac{7}{12}, 0)

Refine

(\frac{7}{12}, 0), (- \frac{7}{12}, 0)

4 (x - 4)^{2} + 9 y^{2} = 36

\frac{( x - 1 ) ^{2}}{45} + \frac{( y - 7 ) ^{2}}{81} = 1

\frac{x ^{2}}{67} + \frac{y ^{2}}{57} = 1

4 x^{2} + 16 y^{2} = 64

\frac{( x + 4 ) ^{2}}{22} + \frac{( y + 3 ) ^{2}}{26} = 1

What is the foci 9x^2+16y^2=1 ?
The foci 9x^2+16y^2=1 is ((sqrt(7))/(12),0),(-(sqrt(7))/(12),0)