Question

Find the locus of the mid-points of the portion of the line x sinθ+ y cosθ = p intercepted between the axes.

Answer 1

We have x sinθ+ y cosθ = p 

\[\Rightarrow \frac{x}{\frac{p}{sin\theta}} + \frac{y}{\frac{p}{cos\theta}} = 1\]

So, the x and y intercepts are given by \[\left( \frac{p}{sin\theta}, 0 \right) \text { and } \left( 0, \frac{p}{cos\theta} \right)\]

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

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

\[ \Rightarrow h = \frac{p}{2sin\theta} \text { and } k = \frac{p}{2cos\theta}\]

\[ \Rightarrow sin\theta = \frac{p}{2h} \text { and } cos\theta = \frac{p}{2k}\]

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

Now, squaring and adding, we get

\[\sin^2 \theta + \cos^2 \theta = \frac{p^2}{4 h^2} + \frac{p^2}{4 k^2}\]

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

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

since, (h, k) is the mid point, so it will also pass through x sinθ+ y cosθ = p. 
Hence, the given equation of locus can also be written as: \[\frac{4}{p^2} = \frac{1}{x^2} + \frac{1}{y^2}\]