Path of the mid-point of AB

Question

Question:

The variable lines goes through the intersection point of $\frac{x}{p}+\frac{y}{q}=1$ and

$\frac{x}{q}+\frac{y}{p}=1$ intersects OX and OY axes at A, B respectively. Show that the path of M which is the mid-point of AB is given by, $2\left(p+q\right)xy=pq(x+y)$ ($p$, $q$ are discrete positive constants)

My attempt:

Any line which goes through the intersection point of above lines can be written, $\left(px+qy-pq\right)+\gamma \left(qx+py-pq\right)=0\ \ \ ;\ \gamma \ is\ a\ parameter$ $\left(p+\gamma q\right)x+\left(\gamma p+q\right)y-pq\left(1+\gamma \right)=0$

$\frac{x}{\frac{pq(1+\gamma )}{\left(p+\gamma q\right)}}+\frac{y} {\frac{pq(1+\gamma )}{\left(\gamma p+q\right)}}=1$

Which implies,

$A \equiv\left(\frac{pq(1+\gamma )}{\left(p+\gamma q\right)},\ 0\right)$

$B \equiv\left(0,\frac{pq(1+\gamma )}{\left(\gamma p+q\right)}\ \right)$

Now, since $M$ is the mid-point of $AB$,

$M \equiv \left(\frac{pq(1+\gamma )}{2\left(p+\gamma q\right)},\frac{pq(1+\gamma )}{2\left(\gamma p+q\right)}\right)$

And since $M$ is variable,

$M\equiv \left(x,y\right)\ \equiv \ \ \left(\frac{pq(1+\gamma )}{2\left(p+\gamma q\right)},\frac{pq(1+\gamma )}{2\left(\gamma p+q\right)}\right)$

$\textit{x} = y\left(\frac{\left(\gamma p+q\right)}{\left(p+\gamma q\right)}\right)$

Here, I am stuck.How do I go about this?

Answer 1

$$x=\frac{pq(1+\gamma )}{2\left(p+\gamma q\right)} \text{ and }y=\frac{pq(1+\gamma )}{2\left(\gamma p+q\right)}$$

Thus, $$\frac{1}{x}+\frac{1}{y}= \frac{2\left(p+\gamma q\right)}{pq(1+\gamma )}+\frac{2\left(\gamma p+q\right)}{pq(1+\gamma )}=\frac{2}{pq(1+\gamma)}(p+q)(1+\gamma)$$ $$=\frac{2(p+q)}{pq}$$ Thus, $$\frac{1}{x}+\frac{1}{y}=\frac{2(p+q)}{pq}$$ Or, $$2\left(p+q\right)xy=pq(x+y)$$