Question

A rectangle $PQRS$ has its side $PQ$ parallel to the line $y=mx$ and vertices $P,Q$ and $S$ on the lines $y=a,x=b$ and $x=-b$, respectively. Find the locus of the vertex $R$.

Answer 1

Point O is the mid point of both line segment PR and SQ so by using section formula for mid point, the x co-ordinates should be same.

$\frac{-b+b}{2}=\frac{J+\alpha }{2}$

$\therefore J=-\alpha$

Line PQ is prallel to line y = mx, so slope of line PQ will be m

$\frac{K-a}{b-J}=m$

$\frac{K-a}{b+\alpha }=m$

$K-a=m(b+\alpha )$

$K=m(b+\alpha )+a⟶Equation\left(1.\right)$

As line SP is perpendicular to line PQ due to property of rectangle, hence slope of line SP should be $\frac{-1}{m}$

$\frac{\beta -K}{\alpha -b}=\frac{-1}{m}$

$\beta -K=\frac{-1}{m}(\alpha -b)$

$K=\beta +\left(\frac{\alpha -b}{m}\right)⟶Equation\left(2.\right)$

Using equation 1 and equation 2, we get

$m(b+\alpha )+a=\beta +\left(\frac{\alpha -b}{m}\right)$

$\left(m\right(b+\alpha )+a)m=\beta m+\alpha -b$

$m^2(b+\alpha )+am=\beta m+\alpha -b$

$m^2\alpha -\alpha -\beta m+b{m}^2+b+am=0$

Now to find locus of point R, replace $\alpha$ with x and $\beta$ with y

we get

$(m^2-1)x-my+b(m^2+1)+am=0$

Which is required locus.