Question

Find the locus of the mid-points of the portion of the line xsinθ+ ycosθ = p intercepted between the axes.

Answer 1

We have xsinθ+ ycosθ = p
$\Rightarrow \frac{x}{{\displaystyle \frac{p}{\mathrm{sin\theta }}}}+\frac{y}{{\displaystyle \frac{p}{\mathrm{cos\theta }}}}=1$
So, the x and y intercepts are given by
$\left(\frac{p}{\mathrm{sin\theta }},0\right)\mathrm{and}\left(0,\frac{p}{\mathrm{cos\theta }}\right)$
Now, let the coordinates of the mid point be (h, k)
$$\begin{gathered} \therefore h=\frac{{\displaystyle \frac{p}{\mathrm{sin\theta }}}+0}{2}\mathrm{and}k=\frac{{\displaystyle 0+\frac{p}{\mathrm{cos\theta }}}}{2} \\ \Rightarrow h=\frac{p}{2\mathrm{sin\theta }}\mathrm{and}k=\frac{p}{2\mathrm{cos\theta }} \\ \Rightarrow \mathrm{sin\theta }=\frac{p}{2h}\mathrm{and}\mathrm{cos\theta }=\frac{p}{2k} \\ \Rightarrow si{n}^2\mathrm{\theta }=\frac{p^2}{4h^2}\mathrm{and}co{s}^2\mathrm{\theta }=\frac{p^2}{4k^2} \end{gathered}$$
Now, squaring and adding, we get
$$\begin{gathered} {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=\frac{p^2}{4h^2}+\frac{p^2}{4k^2} \\ \Rightarrow 1=\frac{p^2}{4h^2}+\frac{p^2}{4k^2} \\ \Rightarrow \frac{4}{p^2}=\frac{1}{h^2}+\frac{1}{k^2} \end{gathered}$$
since, (h, k) is the mid point, so it will also pass through xsinθ+ ycosθ = p.
Hence, the given equation of locus can also be written as:

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