This is a multipart question so bear with me until I get to the part where I am stuck on.

$H$: $xy=c^2$ is a hyperbola

(i) Show that $H$ can be represented by the parametric equations $x=ct$ , $y= \frac{c}{t}$.

If we take $y= \frac{c}{t}$ and rearrange it to $t= \frac{c}{y}$ and subbing this into $x=ct$

$$ x = c(\frac{c}{y}) $$

$$ \therefore xy = c^2 $$

(ii) Find the gradient of the normal to $H$ at the point $T$ with the coordinates $(ct, \frac{c}{t})$

As $xy = c^2$

$$ \Leftrightarrow y = c^2 x^{-1} $$

$$ \Leftrightarrow y' = -\frac{c^2}{x^2} $$

$$ \Leftrightarrow y'_{x=ct} = \frac{-1}{t^2} $$

Hence gradient of the normal is $m = t^2$

The normal to H at the point $T$ meets $H$ again at $P(cp , \frac{c}{p})$

(iiI) By finding the gradient of the line TP and comparing it with the gradient from (ii), find a relationship between $t$ and $p$

So repeating steps of (ii) again

$xy = c^2$

$$ \Leftrightarrow y = c^2 x^{-1} $$

$$ \Leftrightarrow y' = -\frac{c^2}{x^2} $$

$$ \Leftrightarrow y'_{x=cp} = \frac{-1}{p^2} $$

Hence gradient of the normal is $$m = p^2$

Comparing gradients

$$ p^2 = t^2 $$

$$ \Leftrightarrow |p| = |t| $$ Now do we assume $t>0$ and $p>0$? which gives us

$$ \Leftrightarrow p = t $$ as our relationship.

Now this is where I am mostly stuck on

(iv) Hence or otherwise by finding the parametric equations of $M$, the midpoint of $TP$ in terms of $t$, show that the coordinates of $M$ lie on the curve $4x^3y^3 + c^2(x^2-y^2)^2 = 0$

Note: This can also be stated as "Find the Locus of M"

So basically what I have done is to find the normal line of TP

Which is given as

$$ (y-y_1)=m(x-x_1) $$

Because they want everything in terms of $t$ , let $p=t$

$$ (y-\frac{c}{t}) = t^2(x-ct) $$

$$ \Leftrightarrow (y-\frac{c}{t}) = t^2x - ct^3 $$

$$ \Leftrightarrow y = t^2x - ct^3 + \frac{c}{t} $$

Now this is stuck how should I continue?



You have an error here.

That the normal to H at the point $T$ meets $H$ again at $P(cp , \frac{c}{p})$ means that the normal to $H$ at the point $T$ is the line $TP$.

The gradient of the line $TP$ is given by $$\frac{c/p-c/t}{cp-ct}=\frac{-(p-t)}{(p-t)pt}=-\frac{1}{pt}$$ Since this is equal to $t^2$, we have $$-\frac{1}{pt}=t^2\quad\Rightarrow\quad pt^3=-1$$


Here, you have to consider the coordinates of $M$, not the equation of the line $TP$.

Let $M(X,Y)$.

Then, using $p=-1/t^3$, $$X=\frac{cp+ct}{2}=\frac{-c/t^3+ct}{2}=\frac{c(t^4-1)}{2t^3}$$ $$Y=\frac{c/p+c/t}{2}=\frac{-ct^3+c/t}{2}=\frac{c(1-t^4)}{2t}$$ Then, $$\begin{align}&4X^3Y^3+c^2(X^2-Y^2)^2\\&=4\left(\frac{c(t^4-1)}{2t^3}\right)^3\left(\frac{c(1-t^4)}{2t}\right)^3+c^2\left(\left(\frac{c(t^4-1)}{2t^3}\right)^2-\left(\frac{c(1-t^4)}{2t}\right)^2\right)^2\\&=\frac{-c^6(t^4-1)^6}{16t^{12}}+\frac{c^2\cdot c^4(1-t^4)^6}{16t^{12}}\\&=0\end{align}$$


Part (iii): As pointed out by @mathlove, $$p=-1/t^3$$.

Part (iv): Coordinates of $M$ are given by $$x=\frac 12 \left(ct-\frac c{t^3}\right)\qquad\Rightarrow \frac {2x}c=t-\frac 1{t^3}=X\\ y=\frac 12 \left(\frac ct-ct^3\right)\qquad\Rightarrow \frac {2y}c=\frac 1t-t^3=Y\\\\ \frac XY=-\frac 1{t^2}\\ XY=2-t^4-\frac 1{t^4}=-\left(t^2-\frac 1{t^2}\right)^2=-\left(-\frac YX+\frac XY\right)^2\\ (XY)^3=-(X^2-Y^2)^2\\ \left(\frac {2x}c\cdot \frac {2y}c\right)^3=-\left(\left(\frac{2x}c\right)^2-\left(\frac{2y}c\right)^2\right)^2\\ 4x^3y^3=-c^2(x^2-y^2)^2\qquad\blacksquare$$


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