Locus of mid point of the portion between the axes of x cos [ + y, = p] where p is constan…

Question

Locus of mid point of the portion between the axes of

$x$ $cos$ $$\alpha  + y\,\sin \alpha  = p$$  where $p$ is constant is :

Accepted Answer

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 $${1 \over {{x^2}}} + {1 \over {{y^2}}} = {4 \over {{p^2}}}.$$

Solutions · 1