Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

$\alpha$ and $\beta$ satisfy $x\cos \theta +y\sin \theta =2a\Rightarrow \left(x^2+y^2\right){\cos }^2\theta -4ax\cos \theta +\left(4a^2-y^2\right)=0$ So $\cos \alpha +\cos \beta =\frac{4ax}{x^2+y^2}$ and $\cos \alpha \cos \beta =\frac{4a^2-y^2}{x^2+y^2}$. $2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}=1\Rightarrow (1-\cos \alpha )(1-\cos \beta )=1\Rightarrow \cos \alpha +\cos \beta =\cos \alpha \cos \beta$, which gives $4ax=4a^2-y^2$, i.e. $y^2=4a(a-x)$.