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

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


Midpoint of the ladder $(h,k)=(\frac{x-x_1}{2},\frac{x_1}{2})x_1=2k;x=2(h+k)$
We know that,
$(x+x_1{)}^2+(x_1{)}^2=l^2\Rightarrow (2h+4k{)}^2+(2k{)}^2=l^2\Rightarrow \frac{4}{l^2}{x}^2+\frac{16}{l^2}xy+\frac{20}{l^2}{y}^2=1$

Area of an ellipse $A{x}^2+Bxy+C{y}^2=1$
is given by $\frac{2\pi }{\sqrt{4AC-B^2}}$

So area $=\frac{2\pi l^2}{\sqrt{4\times 4\times 5-16}}=\frac{\pi l^2}{4}$