Consider an ellipse with foci at (5, 15) and | vedclass

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(B) The foci of the ellipse are $S'(5, 15)$ and $S(21, 15)$.The distance between the foci is $2ae = \sqrt{(21-5)^2 + (15-15)^2} = 16$,so $ae = 8$.The

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$34$

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(B) The foci of the ellipse are $S'(5, 15)$ and $S(21, 15)$.The distance between the foci is $2ae = \sqrt{(21-5)^2 + (15-15)^2} = 16$,so $ae = 8$.The center of the ellipse is the midpoint of the foci: $(\frac{5+21}{2}, \frac{15+15}{2}) = (13, 15)$.The $X$-axis $(y=0)$ is a tangent to the ellipse. The distance from the center $(13, 15)$ to the tangent line $y=0$ is $15$.For an ellipse,the distance from the center to a tangent line is given by $\sqrt{a^2 \sin^2 	heta + b^2 \cos^2 	heta}$. Since the major axis is horizontal,the tangent is parallel to the major axis,so the distance is simply the semi-minor axis $b = 15$.We know the relation $b^2 = a^2(1 - e^2) = a^2 - (ae)^2$.Substituting the values $b = 15$ and $ae = 8$:$15^2 = a^2 - 8^2$$225 = a^2 - 64$$a^2 = 289$$a = 17$The length of the major axis is $2a = 2 	imes 17 = 34$.

Consider an ellipse with foci at $(5, 15)$ and $(21, 15)$. If the $X$-axis is a tangent to the ellipse,then the length of its major axis equals

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