JEE MAIN - Mathematics (2002 - No. 12)
Locus of mid point of the portion between the axes of
$$x$$ $$cos$$ $$\alpha + y\,\sin \alpha = p$$ where $$p$$ is constant is :
$$x$$ $$cos$$ $$\alpha + y\,\sin \alpha = p$$ where $$p$$ is constant is :
$${x^2} + {y^2} = {4 \over {{p^2}}}$$
$${x^2} + {y^2} = 4{p^2}$$
$${1 \over {{x^2}}} + {1 \over {{y^2}}} = {2 \over {{p^2}}}$$
$${1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}$$
Explanation

Equation of $$AB$$ is
$$x\cos \alpha + y\sin \alpha = p;$$
$$ \Rightarrow {{x\cos \alpha } \over p} + {{y\sin \alpha } \over p} = 1;$$
$$ \Rightarrow {x \over {p/\cos \alpha }} + {y \over {p/\sin \alpha }} = 1$$
So co-ordinates of $$A$$ and $$B$$ are
$$\left( {{p \over {\cos \alpha }},0} \right)$$ and $$\left( {0,{p \over {\sin \alpha }}} \right);$$
So coordinates of midpoint of $$AB$$ are
$$\left( {{p \over {2\cos \,\alpha }},{p \over {2\sin \alpha }}} \right) = \left( {{x_1},{y_1}} \right)\left( {let} \right);$$
$${x_1} = {p \over {2\,\cos \,\alpha }}\,\,\& \,\,{y_1} = {p \over {2\sin \alpha }};$$
$$ \Rightarrow \cos \alpha = p/2{x_1}$$ and $$\sin \alpha = p/2{y_1};$$
$${\cos ^2}\alpha + {\sin ^2}\alpha = 1 \Rightarrow {{{p^2}} \over 4}\left( {{1 \over {{x_1}^2}} + {1 \over {{y_1}^2}}} \right) = 1$$
Locus of $$\left( {{x_1},{y_1}} \right)$$ is $${1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}.$$
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