Question

If $xcos\alpha +ysin\alpha =xcos\beta +ysin\beta =2a$ and $\left[2sin\frac{\alpha }{2}\right]\times sin\frac{\beta }{2}=1$ then :

• A. $y^2=4a(x-a)$
• B. $y^2=4a(a-x)$
• C. $x^2=4a(y-a)$
• D. $x^2=4a(a-y)$

Answer 1

The correct option is B $y^2=4a(a-x)$

Since $\alpha$ and $\beta$ satisfy $x\cos \theta +y\sin \theta =2a$

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

Sum of the roots$=\cos \alpha +\cos \beta =\frac{4ax}{x^2+y^2}$ and $\cos \alpha \cos \beta =\frac{4a^2-y^2}{x^2+y^2}$

Given:$2\sin \frac{\alpha }{2}\sin \frac{\beta }{2}=1$

squaring both sides, we get

$\Rightarrow 4{\sin }^2\frac{\alpha }{2}{\sin }^2\frac{\alpha }{2}=1$

$\Rightarrow \left(2{\sin }^2\frac{\alpha }{2}\right)\left(2{\sin }^2\frac{\beta }{2}\right)=1$

$\Rightarrow (1-\cos \alpha )(1-\cos \beta )=1$ since $1-\cos \theta =2{\sin }^2\frac{\theta }{2}$

$\Rightarrow 1-\cos \alpha -\cos \beta +\cos \alpha \cos beta=1$

$\Rightarrow -\cos \alpha -\cos \beta =-\cos \alpha \cos \beta$

$\Rightarrow \cos \alpha +\cos \beta =\cos \alpha \cos \beta$

$\Rightarrow \frac{4ax}{x^2+y^2}=\frac{4a^2-y^2}{x^2+y^2}$

Multiplying both sides by $x^2+y^2$ we get

$\Rightarrow 4ax=4a^2-y^2$

$\Rightarrow y^2=4a^2-4ax=4a(a-x)$