1. ## Locus of point

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

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 $\frac{a}{b}=\frac{AC}{BC}$
I got
$\frac{a^2}{b^2}=\frac{x^2-4tx+4t^2+y^2}{x^2+y^2-\frac{4y}{t}+\frac{4}{t^2}}$
But now i don't know what to do next, Don't see any ways to make the given equation.

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

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 $\frac{a}{b}=\frac{AC}{BC}$
I got
$\frac{a^2}{b^2}=\frac{x^2-4tx+4t^2+y^2}{x^2+y^2-\frac{4y}{t}+\frac{4}{t^2}}$
But now i don't know what to do next, Don't see any ways to make the given equation.
1. You didn't use the fact that C must be placed on the tangent line with the equation:

$y = -\frac1t^2 x + \frac2t$

2. Sub in the term for y into the equation:

$\frac{a^2}{b^2}=\frac{(x-2t)^2+y^2}{x^2+\left(y-\frac2t\right)^2}$

and solve for x (I've got $x=\frac{2bt}{a+b}~\vee~x=\frac{2bt}{b-a}$)

3. Solve this equation for t and plug in the result into the equation of the tangent line. After expanding the brackets you'll get the given result.

3. I can't seem to get the values for x.

4. ok! got it now! thanks