Find the eccentricity of an ellipse,if the le | vedclass

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. According to the problem,the length of the latus rectum

Vedclass · Accepted Answer

$1/3$

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. According to the problem,the length of the latus rectum is $\frac{2b^2}{a} = 4$,which implies $b^2 = 2a$ ... $(i)$. The distance between the vertex $(a, 0)$ and the nearest focus $(ae, 0)$ is $a - ae = 3/2$,which implies $a(1 - e) = 3/2$ ... $(ii)$. We know that for an ellipse,$b^2 = a^2(1 - e^2)$. Substituting this into equation $(i)$: $\frac{2a^2(1 - e^2)}{a} = 4$ $\Rightarrow 2a(1 - e^2) = 4$ $\Rightarrow a(1 - e^2) = 2$. Since $a(1 - e) = 3/2$,we can write $a(1 - e)(1 + e) = 2$. Substituting $a(1 - e) = 3/2$ into this equation: $\frac{3}{2}(1 + e) = 2$ $\Rightarrow 1 + e = \frac{4}{3}$ $\Rightarrow e = \frac{4}{3} - 1 = \frac{1}{3}$. Thus,the eccentricity is $1/3$.

Find the eccentricity of an ellipse,if the length of its latus rectum is $4$ units and the distance between its vertex and the nearest focus is $3/2$ units.

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