A wall is inclined to the floor at an angle o | vedclass

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Question

(a)

Mid-point $(h, k)=\left(\frac{x-x_1}{2}, \frac{x_1}{2}\right)$

$A B =l$

$A B^2 =l^2$

$\Rightarrow\left(x+x_1\right)^2+x_1^2=l^2$

$\

Vedclass · Accepted Answer

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

Vedclass · Answer

(a)

Mid-point $(h, k)=\left(\frac{x-x_1}{2}, \frac{x_1}{2}ight)$

$A B =l$

$A B^2 =l^2$

$\Rightarrow\left(x+x_1ight)^2+x_1^2=l^2$

$\frac{x-x_1}{2} =h \Rightarrow x-x_1=2 h$

$\frac{x_1}{2} =k \Rightarrow x_1=2 k$

$x+x_1 =2 h+4 k$

Hence, locus of mid-point $(h, k)$ is

$(2 h+4 k)^2+4 k^2 =l^2$

$\Rightarrow \quad x^2+5 y^2+4 x y =\frac{l^2}{4}$

$	ext { Area of ellipse } =\frac{\pi l^2}{4}$

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

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