What is the foci (x^2)/(4^2)+(y^2)/(3^2)=1 ?

The foci (x^2)/(4^2)+(y^2)/(3^2)=1 is (sqrt(7),0),(-sqrt(7),0)

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foci (x^2)/(4^2)+(y^2)/(3^2)=1

foci

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1

(7, 0), (- 7, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4 ^{2}} + \frac{y ^{2}}{3 ^{2}} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 4, b = 3

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

Compute

c :

c = 4^{2} - 3^{2} : 7

(0 + 7, 0), (0 - 7, 0)

Refine

(7, 0), (- 7, 0)

16 x^{2} + 25 y^{2} - 224 x + 250 y = 191

\frac{( x - 1 ) ^{2}}{40} + \frac{( y + 4 ) ^{2}}{36} = 1

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

x^{2} + 4 y^{2} - 6 x + 8 y - 3 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{121} = 1

What is the foci (x^2)/(4^2)+(y^2)/(3^2)=1 ?
The foci (x^2)/(4^2)+(y^2)/(3^2)=1 is (sqrt(7),0),(-sqrt(7),0)