The equation of the ellipse with eccentricity | vedclass

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(B) Given that the eccentricity $e = \frac{1}{2}$ and the foci are at $(\pm ae, 0) = (\pm 1, 0)$.From this,we have $ae = 1$.Substituting $e = \frac{1}

Vedclass · Accepted Answer

$\frac{x^2}{4} + \frac{y^2}{3} = 1$

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(B) Given that the eccentricity $e = \frac{1}{2}$ and the foci are at $(\pm ae, 0) = (\pm 1, 0)$.From this,we have $ae = 1$.Substituting $e = \frac{1}{2}$,we get $a(\frac{1}{2}) = 1$,which implies $a = 2$.Using the relation $b^2 = a^2(1 - e^2)$,we calculate $b^2$:$b^2 = 2^2(1 - (\frac{1}{2})^2) = 4(1 - \frac{1}{4}) = 4(\frac{3}{4}) = 3$.Since the foci are on the $x$-axis,the equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Substituting $a^2 = 4$ and $b^2 = 3$,the equation is $\frac{x^2}{4} + \frac{y^2}{3} = 1$.

The equation of the ellipse with eccentricity $e = \frac{1}{2}$ and foci at $(\pm 1, 0)$ is

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