Question

A wall is inclined to the floor at an angle of ${135}^o$. 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

The correct option is A $\frac{\pi l^2}{4}$

Let the origin be the intersection of wall and ground. Define co-ordinates of the following points :

1) Left end of ladder touching the wall : $(x_l,y_l)$

2) Right and of ladder touching the ground : $(x_r,y_r)$

3) Mid-point of ladder : $(x,y)$

Thus, form the mid-point formula:

$x=\left(\frac{x_l+x_r}{2}\right)$ , $y=\left(\frac{y_l+y_r}{2}\right)$

Since ground lies on the x-axis, $y_r=0$

Wall is at an angle of ${135}^{\circ }\Rightarrow y_l=-x_l$

Length of the ladder is constant. (distance formula)

$\Rightarrow {(x_r-x_l)}^2+{(y_r-y_l)}^2=l^2$

$\Rightarrow {(x_r-x_l)}^2+y_l^2=l^2$ ($y_r=0$) ...(1)

Now,

$y_l=2y$

$x_r-x_l=2(x-x_l)=2(x+2y)$

Therefore, from (1),

$4{(x+2y)}^2+4y^2=l^2$

$\Rightarrow \frac{4x^2}{l^2}+\frac{16xy}{l^2}+\frac{20y^2}{l^2}=1$

( Area of the ellipse $\Rightarrow a{x}^2+bxy+c{y}^2=1$ is given by $\frac{2\pi }{\sqrt{4ac-b^2}}$ )

Area $=\frac{2\pi l^2}{\sqrt{320-256}}$

$=\frac{\pi l^2}{4}$