Search any question & find its solution
Question:
Answered & Verified by Expert
A wall is inclined to the floor at an angle of $135^{\circ}$. A ladder of length $\ell$ is resting on the wall. As the ladder slides down, its mid-point traces an arc of an ellipse. Then the area of the ellipse is
Options:
Solution:
2203 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi \ell^{2}}{4}$
Mid point $(\mathrm{h}, \mathrm{k})=\left(\frac{\mathrm{x}-\mathrm{x}_{1}}{2}, \frac{\mathrm{x}_{1}}{2}\right)$
$\operatorname{Now}\left(\mathrm{x}+\mathrm{x}_{1}\right)^{2}+\mathrm{x}_{1}^{2}=\ell^{2}$
As $2 \mathrm{~h}+4 \mathrm{k}=\mathrm{x}+\mathrm{x}_{1}, 2 \mathrm{k}=\mathrm{x}$,
So required locus is
$$
\begin{array}{l}
4(\mathrm{~h}+2 \mathrm{k})^{2}+4 \mathrm{k}^{2}=\ell^{2} \\
\mathrm{~h}^{2}+5 \mathrm{k}^{2}+4 \mathrm{hk}=\frac{\ell^{2}}{4} \\
\mathrm{x}^{2}+5 \mathrm{y}^{2}+4 \mathrm{xy}=\frac{\ell^{2}}{4}
\end{array}
$$
Whose area is $\frac{\pi \ell^{2}}{4}$
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.