Originally Posted by

**arze** The tangent at the point (t,1/t) on the hyperbola xy=1 meets the x and y axes at A and B respectively. The point C on the line AB is such that AC:AB=a:b. prove that the locus of C as t varies is the rectangular hyperbola

$\displaystyle xy=\frac{4ab}{(a+b)^2}$

I know A(2t,0) and B(0,2/t). So i let C(x,y) and found the values of AC and BC, then used the ratio $\displaystyle \frac{a}{b}=\frac{AC}{BC}$

I got

$\displaystyle \frac{a^2}{b^2}=\frac{x^2-4tx+4t^2+y^2}{x^2+y^2-\frac{4y}{t}+\frac{4}{t^2}}$

But now i don't know what to do next, Don't see any ways to make the given equation.