Question

A wall is inclined to the floor at an angle of $135^{\circ}$. A ladder of length $l$ 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

• A. $\frac{\pi l^{2}}{4}$
• B. $\pi l^{2}$
• C. $4 \pi l^{2}$
• D. $2 \pi l^{2}$

Answer 1

Answer: (A)

Mid-point $\left(h, k\right)=\left(\frac{x-x_{1}}{2}, \frac{x_{1}}{2}\right) $
$AB=l $
$\therefore AB^{2}=l^{2}$
$\Rightarrow \left(x+x_{1}\right)^{2}+x_{1}^{2}=l^{2}$
Now, $\frac{x-x_{1}}{2}=h $
$\Rightarrow x-x_{1}=2h$
$\Rightarrow \frac{x_{1}}{2}=k$
$\Rightarrow x_{1}=2k$
$\therefore x+x_{1}=2h+4k$
Hence, locus of mid-point $\left(h, k\right)$ is
$\left(2h+4k\right)^{2}+4k^{2}=l^{2}$
$\Rightarrow x^{2}+5y^{2}+4xy=\frac{l^{2}}{4}$
$\therefore $ Area of ellipse $=\frac{\pi l^{2}}{4}$