Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1773 Upvotes
Verified Answer
The correct answer is:
$x^2+2 y^2=16$
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.
$\because$ 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.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.