QJEEM02Ma9fbd: JEE Main 2002 Mathematics - Class 11 JEEM02Ma9fbd Solution: Locus of mid point of the portion between the axes of where is…

Question

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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 }}\,\,\&amp; \,\,{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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