+108 The ellipse x^2/49+y^2/b^2=1 has foci at (5,0) and (-5,0).
(a) Fimd the length of a semi major axis of this ellipse
(b)compute the value of b
+1224 The foci of an ellipse are at \((\pm c, 0)\) if the equation of the ellipse is \(\frac{x^2}{49} + \frac{y^2}{b^2} = 1\).
That means that c = 5 and \(b=\sqrt{49-25}=2\sqrt{6}.\)
The length of the semi-major axis is \(2b = 4\sqrt{6}\)
