Question

If $x\cos \alpha +y\sin \alpha =x\cos \beta +y\sin \beta =2a(0<\alpha ,\beta <\pi /2)$, then

• A. $\cos \alpha +\cos \beta =\frac{4ax}{x^2+y^2}$
• B. $\cos \alpha \cos \beta =\frac{4a^2-y^2}{x^2+y^2}$
• C. $\sin \alpha +\sin \beta =\frac{4ay}{x^2+y^2}$
• D. $\sin \alpha +\sin \beta =\frac{4a^2-x^2}{x^2+y^2}$

Answer 1

Answer: (A), (B), (C)

$\cos \alpha \cos \beta =\frac{4a^2-y^2}{x^2+y^2}$
$\sin \alpha +\sin \beta =\frac{4ay}{x^2+y^2}$
$\cos \alpha +\cos \beta =\frac{4ax}{x^2+y^2}$

The general equation becomes

$x\cos \theta +y\sin \theta =2a$.

Or

$(y\sin \theta -2a{)}^2=(-x\cos \theta {)}^2$

Or

$y^2{\sin }^2\theta -4ay\sin \theta +4a^2=x^2{\cos }^2\theta$

Or

$y^2{\sin }^2\theta -4ay\sin \theta +4a^2=x^2(1-{\sin }^2\theta )$

Or

$(x^2+y^2){\sin }^2\theta -4ay\sin \theta +4a^2+x^2=0$.

Let the roots be $\sin \alpha$ and $\sin \beta$.

Hence

$\sin \alpha +\sin \beta =\frac{4ay}{x^2+y^2}$

And

$\sin \alpha .\sin \beta =\frac{4a^2+x^2}{x^2+y^2}$

Similarly

$(x\cos \theta -2a{)}^2=(-y\sin theta{)}^2$

Or

$(x^2+y^2){\cos }^2\theta -4ax\cos \theta +4a^2+y^2=0$

Let the roots be $\cos \alpha$ and $\cos \beta$.

Hence

$\cos \alpha +\cos \beta =\frac{4ax}{x^2+y^2}$

And

$\cos \alpha .\cos \beta =\frac{4a^2+y^2}{x^2+y^2}$