# rectangular hyperbola

• Aug 7th 2009, 05:13 AM
s_ingram
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 x-axis at A and the y-axis 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!
• Aug 7th 2009, 05:34 AM
malaygoel
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)
$\displaystyle xy=\frac{c^2(t^4-1)^2}{4t^4}$....(2)

Sustituting the value of $\displaystyle t^2$ from eq(1) to eq(2)
$\displaystyle xy=\frac{c^2\{(\frac{-y}{x})^2-1\}^2}{4(\frac{-y}{x})^2}$