# Finding the locus - middle point of a line segment

Question: Find the locus of the middle point of the portion of the line $x\cos \alpha + y\sin \alpha = p$ which is intercepted between the axes, given that $p$ remains constant.

Let $(h,k)$ be the middle-point in question. Then \begin{align*} h & = \frac{p}{2\cos \alpha}\\ k & = \frac{p}{2\sin \alpha} \end{align*} Now $\alpha$ is the variable, so we need to eliminate it. Using the fact that $\sin^2 \alpha + \cos^2 \alpha=1$, we get $$\frac{1}{h^2}+\frac{1}{k^2}=\frac{4}{p^2}.$$

Hint:

You can see that the straight line given by your equation intercepts the axes in the points:

$$P\left(0, \frac{p}{\sin{\alpha}} \right), \quad Q\left( \frac{p}{\cos{\alpha}},0 \right).$$

Question:

If you have the coordinates of this two points, how should we find the middle point?

Here's an animation of what is happening here for $\alpha \in [0,\pi/2)$: Hope it helps to visualize the problem.

Cheers!

• The answer is seriously that simple? – Gummy bears Aug 18 '14 at 8:50
• If I dind't misread your question, I'm sure it is! – Dmoreno Aug 18 '14 at 8:51
• Well the guy above answered it I believe. Eliminating $\alpha$ was probably the hardest part :P – Gummy bears Aug 18 '14 at 8:56
• Of course @Gummybears. You can always note that points of the form $\gamma(t) := (\frac{1}{\cos{t}},\frac{1}{\sin{t}})$, $t \in \mathbb{R}$, are a family of hyperbolas. The asymptotes are given by $\sin{t} = 0$ or $\cos{t}=0$. Cheers! – Dmoreno Aug 18 '14 at 8:58

Locus is the set of points that satisfy the given condition.

Here, let $M(h,k)$ is the general point and the given condition is that it must be the midpoint of the distance between the axes of the line $x\cos{\alpha}+y\sin{\alpha}=p$.

We need to find the equation of the locus in terms of $x-$ and $y-$ axis.

$M(h,k)=(\frac{p}{2\cos\alpha},\frac{p}{2\sin\alpha})$

So, $$x^2+y^2=(\frac{p}{2\cos\alpha})^2+(\frac{p}{2\sin\alpha})^2=\frac{p^2}{4.\sin^2\alpha\cos^2\alpha}$$ For $M(h,k)$ we have $x=\frac{p}{2\cos\alpha}$ and $y=\frac{p}{2\sin\alpha}$$\implies \cos\alpha=\frac{p}{2.x} and \sin\alpha=\frac{p}{2y} Substituting,$$ x^2+y^2=\frac{p^2}{4}.\frac{4.y^2.4.x^2}{p^4}=\frac{4.x^2y^2}{p^2}\implies\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$\$