Question

If tangent of any of family of hyperbola $xy=4si{n}^2\theta ,\theta \in (0,2\pi )-\left\{\pi \right\}$ is not a normal of family of circles $x^2+y^2-2x-2y+\mu =0$, where $\mu$ is any real parameter, then set of all values of $\theta$ is:

• A. $(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},\pi )\cup (\pi ,\frac{7\pi }{6})\cup (\frac{11\pi }{6},2\pi )$
• B. $(\pi ,\frac{11\pi }{6})$
• C. $(\frac{\pi }{6},2\pi )$
• D. $(\frac{\pi }{6},\frac{5\pi }{6})\cup (\pi ,\frac{7\pi }{6})\cup (\frac{11\pi }{6},2\pi )$

Answer 1

The correct option is A $(0,\frac{\pi }{6})\cup (\frac{5\pi }{6},\pi )\cup (\pi ,\frac{7\pi }{6})\cup (\frac{11\pi }{6},2\pi )$

Let $y=mx+c$ be a tangent to $xy=4{\sin }^2\theta$.

$\Rightarrow x(mx+c)=4{\sin }^2\theta$ has equal roots.

$\Rightarrow c^2=-16m{\sin }^2\theta$

Now, if the straight line is also a normal to the given circle, then it must pass through the center $(1,1)$.

$\Rightarrow c=1-m$

$\Rightarrow c^2=1+m^2-2m$

$\Rightarrow -16m{\sin }^2\theta =1+m^2-2m$

Hence, if the straight line is not a normal to the circle, then the above quadratic must have non-real roots.
$\Rightarrow D<0,$

$\Rightarrow {(16{\sin }^2\theta -2)}^2<4$

$\Rightarrow {\sin }^2\theta <\frac{1}{4}$

or $-\frac{1}{2}<\sin \theta <\frac{1}{2}$
$\theta \in (0,\frac{\pi }{6})\cup (\frac{5\pi }{6},\pi )\cup (\pi ,\frac{7\pi }{6})\cup (\frac{11\pi }{6},2\pi )$