The length of the latus rectum of an ellipse | vedclass

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Question

(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. The length of the latus rectum is $\frac{2b^2}{a} = 6$,which implies $b^

Vedclass · Accepted Answer

$25 e^2-40 e+16=0$

Vedclass · Answer

(A) Let the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a > b$. The length of the latus rectum is $\frac{2b^2}{a} = 6$,which implies $b^2 = 3a$. The distance between a focus $(ae, 0)$ and its nearest vertex $(a, 0)$ is $a - ae = a(1 - e) = \frac{5}{3}$. We know $b^2 = a^2(1 - e^2)$. Substituting $b^2 = 3a$,we get $3a = a^2(1 - e^2)$,so $3 = a(1 - e^2) = a(1 - e)(1 + e)$. Since $a(1 - e) = \frac{5}{3}$,we have $3 = \frac{5}{3}(1 + e)$. This gives $1 + e = \frac{9}{5}$,so $e = \frac{9}{5} - 1 = \frac{4}{5}$. Now,we check which equation is satisfied by $e = \frac{4}{5}$. For option $A$: $25(\frac{4}{5})^2 - 40(\frac{4}{5}) + 16 = 25(\frac{16}{25}) - 32 + 16 = 16 - 32 + 16 = 0$. Thus,$e$ satisfies $25e^2 - 40e + 16 = 0$.

The length of the latus rectum of an ellipse is $6$ units and the distance between a focus and its nearest vertex on the major axis is $\frac{5}{3}$ units. If $e$ is the eccentricity of this ellipse,then $e$ satisfies the equation:

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