
rectangular hyperbola
Hi folks,
I am struggeling with the following problem:
The tangent at the point P(ct, c/t) (t>0) on the hyperbola $\displaystyle xy = c^2$ meets the xaxis at A and the yaxis at B. The normal at P to the rectangular hyperbola meets the line y = x at C and the line y = x at D.
The normal at P meets the hyperbola again at Q and the mid point of PQ is M. Prove that as t varies, the point M lies on the curve $\displaystyle c^2(x^2  y^2)^2 + 4x^3y^3 = 0$
I have calculated the points as follows: $\displaystyle A(2ct,0), B(0, 2c/t), C(\frac{c(t^2 + 1)}{t}, \frac{c(t^2 + 1)}{t}), D(\frac{c(t^2  1)}{t}, \frac{c(t^2  1)}{t})$
I have the point Q at $\displaystyle (\frac {c}{t^3}, ct^3)$ and the mid point of PQ, point M at $\displaystyle (\frac{c(t^4  1)}{2t^3}, \frac{c(t^4  1)}{2t})$
The locus of M is usually found by eliminating T from the coordinates of the point, but I can't seem to get anywhere. I end up with $\displaystyle t^2 = \frac{y}{x}$ and we have $\displaystyle xy = c^2$ but I can't eliminate t completely. Should I try working back from the given result? I feel there has to be a better way!

you have
$\displaystyle x=\frac{c(t^4  1)}{2t^3}$
$\displaystyle y=\frac{c(t^4  1)}{2t}$
hence,
$\displaystyle \frac{y}{x}=t^2$....(1)
$\displaystyle xy=\frac{c^2(t^41)^2}{4t^4}$....(2)
Sustituting the value of $\displaystyle t^2$ from eq(1) to eq(2)
$\displaystyle xy=\frac{c^2\{(\frac{y}{x})^21\}^2}{4(\frac{y}{x})^2}$