Hi everyone. I can't fully describe things in the title, so here is the full question:

$\displaystyle P(ct, \frac {c}{t})$ is any point on the rectangular hyperbola $\displaystyle xy=c^{2}$. The tangent at P cuts the y-axis at T and the normal at P cuts the x-axis at N. Find the Cartesian equation of the locus of the midpoint of TN.

My attempted solution: First, I define T to be the point $\displaystyle T(0,q)$ and N to be the point $\displaystyle N(p,0)$.

To find the Gradient of the tangent, I implicitly differentiate the hyperbola. Therefore:

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} (xy=c^{2}) $

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = - \frac {y}{x} $

Here, I substitute the $\displaystyle x, y$ values of P into the gradient, and try to solve for PT and PN.

So using this to solve for PT: The gradient of the tangent using the values of P would be $\displaystyle - \frac {1}{t^{2}}$ This makes the equation PT to be:

$\displaystyle (y-0)= - \frac {1}{t^{2}} (x -q) $ Therefore:

$\displaystyle q= t^{2}y+x$

The gradient for the normal would be $\displaystyle t^{2}$, and the equation would be $\displaystyle (y-p)= t^{2} (x -0)$

$\displaystyle p=y-t^{2}x$

Using the found values of p and q, let M be the midpoint of TN. Therefore, $\displaystyle M(X,Y)=(\frac {1}{2} (y-t^{2}x), \frac {1}{2} (t^{2}y+x))$

Here I look like this.

How do I find the gradient to the midpoint so that I can find the Cartesian equation. After being like for some time, I turned over to the answer and it was given as: $\displaystyle y^{4}=c^{2}(c^{2}-2xy)$

What am I doing wrong? Or can c be substituted into the equation later? Can I have some pointers please?

Thanks in advance!!