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10-2. Parabola, Ellipse, Hyperbola
normal
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
$2 \pi l^2$
(KVPY-2016)
Solution
(a)
Mid-point $(h, k)=\left(\frac{x-x_1}{2}, \frac{x_1}{2}\right)$
$A B =l$
$A B^2 =l^2$
$\Rightarrow\left(x+x_1\right)^2+x_1^2=l^2$
$\frac{x-x_1}{2} =h \Rightarrow x-x_1=2 h$
$\frac{x_1}{2} =k \Rightarrow x_1=2 k$
$x+x_1 =2 h+4 k$
Hence, locus of mid-point $(h, k)$ is
$(2 h+4 k)^2+4 k^2 =l^2$
$\Rightarrow \quad x^2+5 y^2+4 x y =\frac{l^2}{4}$
$\text { Area of ellipse } =\frac{\pi l^2}{4}$
Standard 11
Mathematics
