A line of fixed length $$2$$ units moves so that its ends are on the positive $$x-axis$$ and that part of the line x $$+$$ y $$=$$ $$0$$ which lies in the second quadrant. Then the locus of the $$mid-point$$ of the line has the equation

Question

A line of fixed length $$2$$ units moves so that its  ends are on the positive $$x-axis$$ and that part of the line x $$+$$ y $$=$$ $$0$$ which  lies in the second quadrant. Then the locus of the $$mid-point$$ of the line has  the equation

Infinity Learn · Accepted Answer

If ∠$$BAO=$$θ  then  $$BM=2sin$$θ  and $$MO=BM=2sin$$θ,$$MA=2cos$$θ .                     Hence $$A=(2cos$$θ−$$2sin$$θ,$$0)$$  and $$B($$−$$2sin$$θ,$$2sin$$θ$$)$$                                                              Since P $$($$ x, $$y)$$ is the mid point of AB ,                                                                                             $$2x=(2cos$$θ$$)+($$−$$4sin$$θ$$)$$  or  $$cos$$θ−$$2sin$$θ$$=x$$                                                                          $$2y=(2sin$$θ$$)$$  or  $$sin$$θ$$=y$$                                                                                                                                                                                                  Eliminating θ , we have   $$(x+2y)2+y2=1$$  or ,$$x2+5y2+4xy$$−$$1=0$$

A line of fixed length 2 units moves so that its ends are on the positive x-axis and that part of the line x + y = 0 which lies in the second quadrant. Then the locus of the mid-point of the line has the equation

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