If the normal to the rectangular hyperbola x 2 - y 2 = 1 at the point P π 4 meets the curve again at Q…

Given,

Equation of hyperbola $x^2-y^2=1$

Now parametric point is given by $\left(a\text{sec}\theta ,b\tan \theta \right)\equiv \left(\sqrt{2},1\right)$ as given $\theta =45° \& a=b=1$

Now finding slope of tangent at the given point by differentiating the curve we get,

 $2x-2y\frac{dy}{dx}=0$

$\Rightarrow {\left(\frac{dy}{dx}\right)}_{\left(\sqrt{2},1\right)}=\frac{\sqrt{2}}{1}$

So, slope of normal will be $-\sqrt{2}$,

Now equation of normal will be $y=\frac{-x}{\sqrt{2}}+2$

Since normal is intersecting the curve again so,

$x^2-{\left(\frac{-x}{\sqrt{2}}+2\right)}^2=1$

$\Rightarrow x^2+4\sqrt{2}x-10=0$

$\Rightarrow x=\sqrt{2} \text{or }-5\sqrt{2}$

So, $y=7$ for $x=-5\sqrt{2}$

Now we know that $x=\text{sec}\theta =-5\sqrt{2}$

And $y=\tan \theta =7$,

So, the value of ${\text{sec}}^2\theta +\tan \theta =50+7=57$