The ellipse having its foci (0, pm 1) and maj | vedclass

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Question

(A) Since the foci of the given ellipse lie on the $Y$-axis,it is a vertical ellipse. Let the required equation be $\frac{x^2}{b^2} + \frac{y^2}{a^2}

Vedclass · Accepted Answer

$20 x^2+4 y^2=5$

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(A) Since the foci of the given ellipse lie on the $Y$-axis,it is a vertical ellipse. Let the required equation be $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$,where $a^2 > b^2$. The foci are $(0, \pm c) = (0, \pm 1)$,which implies $c = 1$. The length of the major axis is $2a = \sqrt{5}$,so $a = \frac{\sqrt{5}}{2}$ and $a^2 = \frac{5}{4}$. Using the relation $c^2 = a^2 - b^2$,we have $1 = \frac{5}{4} - b^2$. Thus,$b^2 = \frac{5}{4} - 1 = \frac{1}{4}$. Substituting $a^2$ and $b^2$ into the equation,we get $\frac{x^2}{1/4} + \frac{y^2}{5/4} = 1$. This simplifies to $4x^2 + \frac{4y^2}{5} = 1$,or $20x^2 + 4y^2 = 5$.

The ellipse having its foci $(0, \pm 1)$ and major axis of length $\sqrt{5}$ is

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