Question

$P$ is a variable point on the line $y=4$. Tangents are drawn to the circle $x^2+y^2=4$ from $P$ to touch it at $A$ and $B$. The parallelogram $PAQB$ is completed. Prove that the locus of the point $Q$ is $(y+4)(x^2+y^2)=2y^2$.

Answer 1

$P$ lies on $y=4$ and hence its co-ordinates be taken as $(h,4)$, then $AB$ is the chord of contact with respect to circle $x^2+y^2=4$ whose equation is
$hx+4y=4....\left(1\right)$
Solving with circle, we get
$x^2+{\left(\frac{4-hx}{4}\right)}^2=4$
or $x^2(16+h{)}^2-8hx-48=0$
Above gives abscissas of the points $A$ and $B$
$\therefore x_1+x_2=\frac{8h}{16+h^2}$.
Also the points $A$ and $B$ lie on $\left(1\right)$
$\therefore 4(y_1+y_2)=8-\frac{h.8h}{16+h^2}$
$\therefore y_1+y_2=\frac{32}{16+h^2}$
Now if the point $Q$ be $(\alpha ,\beta )$, then the figure $PAQB$ being a parallelogram its diagonals bisect
$\therefore x_1+x_2=h+\alpha =\frac{8h}{16+h^2}.....\left(2\right)$
$y_1+y_2=4+\beta =\frac{32}{16+h^2}.....\left(3\right)$
Now we have to eliminate the variable between $\left(2\right)$ and $\left(3\right)$ to find the locus of $Q$, i.e. $(\alpha ,\beta )$.
Dividing $\frac{h+\alpha }{4+\beta }=\frac{h}{4}\therefore 4\alpha =h\beta$
or $h=\frac{4\alpha }{\beta }$. Put in $\left(3\right)$ and we get
$(4+\beta )(16+\frac{16{\alpha }^2}{{\beta }^2})=32$
$\therefore$ Locus is $(y+4)(x^2+y^2{)}_{=}2y^2$.