Question

Let $\theta$ be a variable angle, find the locus of the point (x, y) when $x=a\tan (\theta +\alpha )$ and $y=b\tan (\theta +\beta )$

Answer 1

Given: $x=a\tan (\theta +\alpha )$ and $y=b\tan (\theta +\beta )$

To get the required locus, we are supposed to eliminate 't' from these equations.

$\Rightarrow$ $x=a\tan (\theta +\alpha )\Rightarrow \theta ={\tan }^{-1}\left(\frac{x}{a}\right)-\alpha ...eqn\left(i\right)$

$\Rightarrow$ $y=b\tan (\theta +\beta )\Rightarrow \theta ={\tan }^{-1}\left(\frac{y}{b}\right)-\beta ...eqn\left(ii\right)$

Equating the eqns (i) & (ii), we get

$\Rightarrow$ ${\tan }^{-1}\left(\frac{x}{a}\right)-\alpha$ $=$ ${\tan }^{-1}\left(\frac{y}{b}\right)-\beta$

$\Rightarrow$ ${\tan }^{-1}\left(\frac{x}{a}\right)-{\tan }^{-1}\left(\frac{y}{b}\right)$ $=$ $\alpha -\beta$

Apply ${\tan }^{-1}P-{\tan }^{-1}Q={\tan }^{-1}\left(\frac{P-Q}{1+PQ}\right)$

$\Rightarrow$ $\frac{\frac{x}{a}-\frac{y}{b}}{1+\frac{x}{a}.\frac{y}{b}}=\tan (\alpha -\beta )$

$\Rightarrow$ $(bx-ay)=\tan (\alpha -\beta )(ab+xy)$

This is the required locus of the point (x,y).