What is the foci ((x+20)^2)/(105)+(y^2)/(16)=1 ?

The foci ((x+20)^2)/(105)+(y^2)/(16)=1 is (-20+sqrt(89),0),(-20-sqrt(89),0)

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foci ((x+20)^2)/(105)+(y^2)/(16)=1

foci

(- 20 + 89, 0), (- 20 - 89, 0)

Solution steps

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

Calculate ellipse properties

\frac{( x + 20 ) ^{2}}{105} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (- 20, 0), a = 105, b = 4

(- 20 + c, 0), (- 20 - c, 0)

Compute

c :

c = 105^{2} - 4^{2} : 89

(- 20 + 89, 0), (- 20 - 89, 0)

What is the foci ((x+20)^2)/(105)+(y^2)/(16)=1 ?
The foci ((x+20)^2)/(105)+(y^2)/(16)=1 is (-20+sqrt(89),0),(-20-sqrt(89),0)