Question

Find the locus of the mid-points of the portion of the line $xsin\theta +ycos\theta =p$ intercipted between the axes.

Answer 1

We have $xsin\theta +ycos\theta =p$

$\Rightarrow \frac{x}{\frac{p}{sin\theta }}+\frac{y}{\frac{p}{sin\theta }}=1$

So, the x and y intercepts are given by

$(\frac{p}{sin\theta },0)$ and $(0,\frac{p}{cos\theta })$

Now, let coordinates of the mid point be (h,k)

$\therefore h=\frac{\frac{p}{sin\theta }+0}{2}$ and $k=\frac{0+\frac{p}{cos\theta }}{2}$

$\Rightarrow h=\frac{p}{2sin\theta }$ and $k=\frac{p}{2cos\theta }$

$\Rightarrow sin\theta =\frac{p}{2h}$ and $cos\theta =\frac{p}{2k}$

$\Rightarrow si{n}^2\theta =\frac{p^2}{4h^2}$ and $co{s}^2\theta =\frac{p^2}{4k^2}$

Now,squaring and adding, we get

$si{n}^2\theta +co{s}^2\theta =\frac{p^2}{4h^2}+\frac{p^2}{4k^2}$

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

$\Rightarrow \frac{4}{4h^2}+\frac{1}{h^2}+\frac{1}{k^2}$

Since, (h,k) is the mid point, so it will also pass through$xsin\theta +ycos\theta =p.$ Hence,the given equation of locus can also be written as :

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