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 $\left(h,k\right)=\left(\frac{x-x_1}{2},\frac{x_1}{2}\right)$
$AB=l$
$\therefore A{B}^2=l^2$
$\Rightarrow {\left(x+x_1\right)}^2+x_1^2=l^2$
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 $=\frac{\pi l^2}{4}$