The major and minor axes of an ellipse are along the $X$-axis and $Y$-axis respectively. If its latusrectum is of length 4 and the distance between the foci is $4 \sqrt{2}$, then the equation of that ellipse is

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Aicharya · Accepted Answer

Let the equation of ellipse

$\because$ Length of latusrectum of ellipse (i) is

and distance between the foci is $ \begin{array}{rlr} & 2 a e=4 \sqrt{2} & \\ \Rightarrow & a \sqrt{1-\frac{b^2}{a^2}}=2 \sqrt{2} & {\left[\because e=\sqrt{1-\frac{b^2}{a^2}},(b>a)\right]} \\ \Rightarrow & a^2-b^2=8 & \\ \Rightarrow & a^2-2 a-8=0 & \text { [from Eq. (ii) } \left., b^2=2 a\right] \\ \Rightarrow & (a-4)(a+2)=0 & \\ \Rightarrow & a=4,-2 & \\ \Rightarrow & a=4 & \{\because a>0\} \end{array} $ So, $b^2=8$ $\therefore$ Equation of required ellipse is $ \begin{aligned} & \frac{x^2}{16}+\frac{y^2}{8}=1 \\ \Rightarrow \quad x^2+2 y^2 & =16 \end{aligned} $ Hence, option (b) is correct

The major and minor axes of an ellipse are along the $X$-axis and $Y$-axis respectively. If its latusrectum…

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