Question

If $(x-a)\cos \theta +y\sin \theta =(x-a)\cos \varphi +y\sin \varphi =a$ and $\tan \left(\theta /2\right)-\tan \left(\varphi /2\right)=2b,$ then

• A. $\tan \frac{\varphi }{2}=\frac{1}{x}(y-bx)$
• B. $y^2=2bx-(1-a^2)x$
• C. $\tan \frac{\theta }{2}=\frac{1}{x}(y+bx)$
• D. $y^2=2ax-(1-b^2)x^2$

Answer 1

Answer: (A), (C), (D)

$y^2=2ax-(1-b^2)x^2$

Let $\tan \left(\theta /2\right)=\alpha$ and $\tan (\varphi /2=\beta )$
$\Rightarrow \alpha -\beta =2b$
Also $\cos \theta =\frac{1-{\alpha }^2}{1+{\alpha }^2}$ at $\sin \theta =\frac{2\alpha }{1+{\alpha }^2}$
Similarly $\cos \varphi =\frac{1-{\beta }^2}{1+{\beta }^2};\sin \varphi =\frac{2\beta }{1+{\beta }^2}$
Therefore, we have from the given relations
$(x-a)\frac{1-{\alpha }^2}{1+{\alpha }^2}+y\left(\frac{2\alpha }{1+{\alpha }^2}\right)=a$
$\Rightarrow x{\alpha }^2-2y\alpha +2a-x=0$ and $x{\beta }^2-2y\beta +2a-x=0$
Similarly we see that $\alpha$ and $\beta$ are the roots of the equation $x{z}^2-2yz+2a-x=0,$ so that $\alpha +\beta =2y/x$ and $\alpha \beta =\frac{(2a-x)}{x}$.
Now, from ${(\alpha +\beta )}^2={(\alpha -\beta )}^2+4\alpha \beta ,$ we get
${\left(\frac{2y}{x}\right)}^2={\left(2b\right)}^2+\frac{4(2a-x)}{x}$
$\Rightarrow y^2=2ax-(1-b^2)x^2$ and $\alpha +\beta =\frac{2y}{x}$ and $\alpha -\beta =2b$
$\Rightarrow \alpha =\frac{y}{x}+b,\beta =\frac{y}{x}-b$
$\Rightarrow \frac{\tan \theta }{2}=\frac{1}{x}(y+bx)$ and $\frac{\tan \varphi }{2}=\frac{1}{x}(y-bx)$