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The value of a for the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ if the extremities of the latusrectum of the ellipse having positive
ordinates lie on the parabola $x^{2} = - 2 (y - 2)$ is

Moderate

Solution

$(\pm ae, \frac{b^{2}}{a})$ are extremities of the latusrectum having positive ordinates. Then,
$\begin{aligned} a^{2} e^{2} = - 2 (\frac{b^{2}}{a} - 2) - - - - (1) \\ But b^{2} = a^{2} (1 - e^{2}) - - - - (2) \end{aligned}$
Therefore, from (1) and (2), we get
$\begin{array}{l} a^{2} e^{2} - 2 {ae}^{2} + 2 a - 4 = 0 \\ or {ae}^{2} (a - 2) + 2 (a - 2) = 0 \\ ∴ ({ae}^{2} + 2) (a - 2) = 0 \end{array}$
Hence, a=2.

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Ellipse

Remember concepts with our Masterclasses.

The value of a for the ellipse x2a2+y2b2=1,(a>b) if the extremities of the latusrectum of the ellipse having positiveordinates lie on the parabola x2=−2(y−2) is

±ae,b2a are extremities of the latusrectum having positive ordinates. Then,a2e2=−2b2a−2----1But b2=a21−e2----2Therefore, from (1) and (2), we geta2e2−2ae2+2a−4=0or ae2(a−2)+2(a−2)=0∴ae2+2(a−2)=0Hence, a=2.

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The value of a for the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ if the extremities of the latusrectum of the ellipse having positive
ordinates lie on the parabola $x^{2} = - 2 (y - 2)$ is