2
$\begingroup$

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.

No idea how to even approach this problem. Please help!

$\endgroup$
2
$\begingroup$

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 $$\frac{1}{h^2}+\frac{1}{k^2}=\frac{4}{p^2}.$$

$\endgroup$
0
$\begingroup$

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?


Addendum:

Here's an animation of what is happening here for $\alpha \in [0,\pi/2)$:

enter image description here

Hope it helps to visualize the problem.

Cheers!

$\endgroup$
  • $\begingroup$ The answer is seriously that simple? $\endgroup$ – Gummy bears Aug 18 '14 at 8:50
  • $\begingroup$ If I dind't misread your question, I'm sure it is! $\endgroup$ – Dmoreno Aug 18 '14 at 8:51
  • $\begingroup$ Well the guy above answered it I believe. Eliminating $\alpha$ was probably the hardest part :P $\endgroup$ – Gummy bears Aug 18 '14 at 8:56
  • $\begingroup$ 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! $\endgroup$ – Dmoreno Aug 18 '14 at 8:58
0
$\begingroup$

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,

$$ x^2+y^2=\frac{p^2}{4}.\frac{4.y^2.4.x^2}{p^4}=\frac{4.x^2y^2}{p^2}\implies\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.