Question

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

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

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\tan \frac{\theta }{2}=\frac{1}{x}(y+bx)$
B $y^2=2ax-(1-b^2)x^2$
D $\tan \frac{\varphi }{2}=\frac{1}{x}(y-bx)$

Let,

$\tan \left(\frac{\theta }{2}\right)=\alpha$ and $\tan \left(\frac{\varphi }{2}\right)=\beta$

So that, $\alpha -\beta =2b$

Also, $\cos \theta =\frac{1-{\tan }^2\left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}=\frac{1-{\alpha }^2}{1+{\alpha }^2}$

and $\sin \theta =\frac{2\tan \left(\frac{\theta }{2}\right)}{1+{\tan }^2\left(\frac{\theta }{2}\right)}=\frac{2\alpha }{1+{\alpha }^2}$

similarly, $\cos \varphi =\frac{1-{\beta }^2}{1+{\beta }^2}$ and $\sin \varphi =\frac{2\varphi }{1+{\beta }^2}$

Therefore, we have from the given relation, $(x-a)\frac{1-{\alpha }^2}{1+{\alpha }^2}+y\left(\frac{2\alpha }{1+{\alpha }^2}\right)=a$

$\Rightarrow x{\alpha }^2-2y^2+2a-x=0$

similarly, $\Rightarrow x{\beta }^2-2y\beta +2a-x=0$

We see that $\alpha$ and $\beta$ are some roots of the equation
$x{z}^2-2yz+2a-x=0$

So that $\alpha +\beta =\frac{2y}{x}$ and $\alpha \beta =\frac{(2a-x)}{x}$

Now from
${(\alpha +\beta )}^2={(\alpha -\beta )}^2+4\alpha \beta$

$\Rightarrow {\left(\frac{2y}{x}\right)}^2={\left(2b\right)}^2+\frac{4(2a-x)}{x}\Rightarrow y^2-2ax-(1-b^2)x$

Also from $\alpha +\beta =\frac{2y}{x},\alpha -\beta =2b$
we get $\alpha =\frac{y}{x}+b$ and $\beta =\frac{y}{x}-b$

$\tan \left(\frac{\theta }{2}\right)=\frac{1}{x}(y+bx)$ and $\tan \left(\frac{\varphi }{2}\right)=\frac{1}{x}(y-bx)$