Let E be an ellipse whose major axis is the X | vedclass

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(A) By the definition of an ellipse,the sum of the distances of any point on the ellipse from its two foci is equal to the length of the major axis,$2

Vedclass · Accepted Answer

$3/5$

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(A) By the definition of an ellipse,the sum of the distances of any point on the ellipse from its two foci is equal to the length of the major axis,$2a$. Given $SP_1 + SP_2 = 2a$,where $SP_1 = \frac{7}{2}$ and $SP_2 = \frac{13}{2}$. $2a = \frac{7}{2} + \frac{13}{2} = \frac{20}{2} = 10 \Rightarrow a = 5$. The foci are at $(\pm ae, 0) = (\pm 5e, 0)$. Let $S = (5e, 0)$ and $P = \left(\frac{5}{2}, 2\sqrt{3}\right)$. The distance $SP = \frac{7}{2}$. $SP^2 = \left(5e - \frac{5}{2}\right)^2 + (2\sqrt{3} - 0)^2 = \left(\frac{7}{2}\right)^2$. $\left(5e - \frac{5}{2}\right)^2 + 12 = \frac{49}{4}$. $\left(5e - \frac{5}{2}\right)^2 = \frac{49}{4} - 12 = \frac{49 - 48}{4} = \frac{1}{4}$. $5e - \frac{5}{2} = \pm \frac{1}{2}$. Case $1$: $5e = \frac{5}{2} + \frac{1}{2} = 3 \Rightarrow e = \frac{3}{5}$. Case $2$: $5e = \frac{5}{2} - \frac{1}{2} = 2 \Rightarrow e = \frac{2}{5}$. Since $e = 3/5$ is one of the options,the eccentricity is $3/5$.

Let $E$ be an ellipse whose major axis is the $X$-axis and minor axis is the $Y$-axis. If the distances of a point $P \left(\frac{5}{2}, 2 \sqrt{3}\right)$ on $E$ from its foci are $\frac{7}{2}$ and $\frac{13}{2}$,then the eccentricity of the ellipse $E$ is

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