Question

The number of normal(s) of a rectangular hyperbola which can touch its conjugate is equal to

Answer 1

In the question it is asked to find the number of normals to a rectanguar hyperbola, which is tangent to its conjugate hyperbola.
Now, let's suppose hyperbola to be $xy=c^{2}$ . So, its conjugate hyperbola will be $xy=-c^{2}$ .
Now, normal to the hyperbola $xy=c^{2}$ at the parametric point $\left(c t , \frac{c}{t}\right)$ is given by $y-\frac{c}{t}=t^{2}\left(x - c t\right)$ .
Since, it is tangent to conjugate hyperbola, on solving with conjugate hyperbola $xy=-c^{2}$ , we should be getting equal roots of formed quadratic equation.
Now, on substituting the value of $y$ from equation of tangent to hyperbola we get, $x\left\{\frac{c}{t} + t^{2} \left(x - c t\right)\right\}+c^{2}=0$
$\Rightarrow t^{2}x^{2}+\left\{\frac{c}{t} - c t^{3}\right\}x+c^{2}=0$
Since the above equation has equal roots, $D=0$ .
$\Rightarrow \left(\frac{1}{t} - t^{3}\right)^{2}-4t^{2}=0$
$\Rightarrow \left(1 - t^{4}\right)^{2}-4t^{4}=0$
$\Rightarrow t^{4}-2t^{2}-1=0$ or $t^{4}+2t^{2}-1=0$
$\Rightarrow t^{2}=\frac{2 \pm \sqrt{8}}{2}$ or $\frac{- 2 \pm \sqrt{8}}{2}$
$\Rightarrow t^{2}=1\pm\sqrt{2}$ or $-1\pm\sqrt{2}$
Reject negative values
$\Rightarrow t^{2}=1+\sqrt{2}$ or $-1+\sqrt{2}$
Since for $t^{2}$ we are getting two values.
Thus, total four values of $t$ exists.