# Thread: Finding the Midpoint of the line connecting two points which are...

1. ## Finding the Midpoint of the line connecting two points which are...

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

$P(ct, \frac {c}{t})$ is any point on the rectangular hyperbola $xy=c^{2}$. 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 $T(0,q)$ and N to be the point $N(p,0)$.

To find the Gradient of the tangent, I implicitly differentiate the hyperbola. Therefore:
$\frac{\mathrm{d} y}{\mathrm{d} x} (xy=c^{2})$

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

Here, I substitute the $x, y$ 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 $- \frac {1}{t^{2}}$ This makes the equation PT to be:

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

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

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

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

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: $y^{4}=c^{2}(c^{2}-2xy)$

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

2. ## Re: Finding the Midpoint of the line connecting two points which are...

I suspect that my differentiation is incorrect, but I can't find the correct differentiation.

3. ## Re: Finding the Midpoint of the line connecting two points which are...

I am having a little trouble understanding your notation. The notation $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ means the derivative of $y$ with respect to $x$. The notation $\dfrac{\mathrm{d}y}{\mathrm{d}x}(xy=c^2)$ doesn't make sense. Am I correct in assuming you mean implicitly differentiate both sides of the equation $xy=c^2$ with respect to $x$? If that is the case, the correct notation would be $\dfrac{\mathrm{d}}{\mathrm{d}x}(xy) = \dfrac{\mathrm{d}}{\mathrm{d}x}(c^2)$ which reads the derivative of $xy$ with respect to $x$ is equal to the derivative of $c^2$ with respect to $x$.

Anyway, assuming my understanding of your notation is correct, then the equation you are using for the tangent and normal lines are wrong. The tangent line at $\left(ct,ct^{-1}\right)$ is given by the equation $y-ct^{-1} = -t^{-2}(x-ct)$. If $x=0$, then you will get a point on the $y$-axis. So, let $x=0$ and solve for $y$:

$y = ct^{-1} + ct^{-1} = \dfrac{2c}{t}$

So $T = \left(0,\dfrac{2c}{t} \right)$.

Next, the equation of the normal line is given by $y-ct^{-1} = t^2(x-ct)$. To find the $x$-intercept, let $y=0$ and solve for $x$:

$x = -ct^{-3}+ct$

So $N = \left(c\dfrac{t^4-1}{t^3}, 0 \right)$.

Now, the midpoint is $\left( c\dfrac{t^4-1}{2t^3}, \dfrac{c}{t} \right)$.

4. ## Re: Finding the Midpoint of the line connecting two points which are...

Thanks a lot. Like I thought, my derivative was wrong. Yes, I was speaking of implicit differentiation. I think I pressed the wrong button in the editor. (I use a LaTeX editor).

From your workings, it looks like I don't need to use differentiation in my question at all after all. Am I correct?

5. ## Re: Finding the Midpoint of the line connecting two points which are...

Actually, you calculated the derivative correctly. I did not rewrite it, but in order to get the slope of the tangent and normal lines, I used the derivative you obtained.

$\left. \dfrac{\mathrm{d}y}{\mathrm{d}x}\right]_{\left(ct,ct^{-1}\right)} = -\dfrac{y}{x} = -\dfrac{ct^{-1}}{ct} = -\dfrac{1}{t^2}$

where the notation $\left. \dfrac{\mathrm{d}y}{\mathrm{d}x}\right]_{\left(ct,ct^{-1}\right)}$ means the evaluation of the derivative at the point $(ct,ct^{-1})$.

So, the slope of the tangent line is $-\dfrac{1}{t^2}$ and the slope of the normal line is $t^2$ (the slope of the normal line is negative the reciprocal of the slope of the tangent line).

So, everything else was your work. The only difference between our two results is that rather than writing the equations of the lines using the unknown points $T,N$, I just wrote the equations using the same known point $(ct,ct^{-1})$. Everything else was your work.

Do you see how to get from the point $\left(c\dfrac{t^4-1}{2t^3},\dfrac{c}{t}\right)$ to the equation $y^4 = c^2(c^2-2xy)$?

6. ## Re: Finding the Midpoint of the line connecting two points which are...

The Gradient of M is $\frac {y_T-y_N}{x_T-x_N}= \frac {(\frac {2c}{t})}{-\frac {c(t^{4}-1)}{t^{3}}}$

Simplifying, this would be $-\frac {2t^{2}}{t^{4}-1}$

Inputting then $y-\frac {c}{t}= (-\frac {2t^{2}}{t^{4}-1})(x-\frac {c(t^{4}-1)}{t^{3}})$

Simplifying, I get $t^{5}(y)-t^{4}(3c)+t^{3}(2x)+t(c-y)+2=0$

I don't know how to get rid of the variable t.

Help on that would be appreciated!

7. ## Re: Finding the Midpoint of the line connecting two points which are...

Originally Posted by LimpSpider
The Gradient of M is $\frac {y_T-y_N}{x_T-x_N}= \frac {(\frac {2c}{t})}{-\frac {c(t^{4}-1)}{t^{3}}}$

Simplifying, this would be $-\frac {2t^{2}}{t^{4}-1}$

Inputting then $y-\frac {c}{t}= (-\frac {2t^{2}}{t^{4}-1})(x-\frac {c(t^{4}-1)}{t^{3}})$

Simplifying, I get $t^{5}(y)-t^{4}(3c)+t^{3}(2x)+t(c-y)+2=0$

I don't know how to get rid of the variable t.

Help on that would be appreciated!
You should not be taking any gradients. You are using the midpoint $(x,y) = \left(\dfrac{c(t^4-1)}{2t^3}, \dfrac{c}{t}\right)$, so $x = \dfrac{c(t^4-1)}{2t^3}, y = \dfrac{c}{t}$. Solve for $t$ in the second equation and plug it into the first:

\begin{align*}t & = \dfrac{c}{y} \\ x & = \dfrac{c\left[\left(\tfrac{c}{y}\right)^4-1\right]}{2\left(\tfrac{c}{y}\right)^3}\end{align*}

Multiply top and bottom of the RHS by $c\left(\dfrac{y}{c}\right)^4$:
$x = \dfrac{c^2\left(1 - \tfrac{y^4}{c^4}\right)}{2y}$

Multiply both sides by 2y:
\begin{align*}2xy & = c^2 - \dfrac{y^4}{c^2} \\ \dfrac{y^4}{c^2} & = c^2 - 2xy \\ y^4 & = c^2(c^2-2xy)\end{align*}

8. ## Re: Finding the Midpoint of the line connecting two points which are...

Thanks! That helps!