Question

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

Answer 1

Answer: 4

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 $\left(ct,\frac{c}{t}\right)$ is given by $y-\frac{c}{t}=t^2\left(x-ct\right)$ .
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, $x\left\{\frac{c}{t}+t^2\left(x-ct\right)\right\}+c^2=0$
$\Rightarrow t^2{x}^2+\left\{\frac{c}{t}-c{t}^3\right\}x+c^2=0$
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.