Question

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

Answer 1

$x=a\cos (\theta +\alpha ).....\left(i\right)y=b\cos (\theta +\beta ).....\left(ii\right)2xy=ab2\cos (\theta +\alpha )\cos (\theta +\beta )$

Using $\cos \left(C\right)+\cos \left(D\right)=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

$\frac{2xy}{ab}=\cos (\theta +\alpha +\theta +\beta )+\cos (\theta +\alpha -\theta -\beta )\frac{2xy}{ab}=\cos (\theta +\theta +\alpha +\beta )+\cos (\alpha -\beta )\cos (\theta +\theta +\alpha +\beta )=\frac{2xy}{ab}-\cos (\alpha -\beta )........\left(iii\right)$

using $\left(i\right)$ and $\left(ii\right)$

${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2={\cos }^2(\theta +\alpha )+{\cos }^2(\theta +\beta ){\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2={\cos }^2(\theta +\alpha )-{\sin }^2(\theta +\beta )+1$

Using ${\cos }^{2}A-{\sin }^{2}B=\cos (A-B)\cos (A+B)$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos (\theta +\alpha +\theta +\beta )\cos (\theta +\alpha -\theta -\beta )+1\frac{x^2}{a^2}+\frac{y^2}{b^2}=\cos (\theta +\theta +\alpha +\beta \left)\cos (\alpha -\beta \right)+1$

Substituting$\left(iii\right)$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=(\frac{2xy}{ab}-\cos (\alpha -\beta ))\cos (\alpha -\beta )+1\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos (\alpha -\beta )=1-{\cos }^2(\alpha -\beta )\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{2xy}{ab}\cos (\alpha -\beta )={\sin }^2(\alpha -\beta )$

is the required locus of $(x,y)$