Question

The number of integral values of exhaustive set of $a^2$ such that there exists a tangent to the ellipse $x^2+a^2{y}^2=a^2$ and the portion of the tangent intercepted by the hyperbola $a^2{x}^2-y^2=1$ subtends a right angle at the centre of the curves is

Answer 1

The equation of tangent at point $(a\cos \theta ,\sin \theta )$ on the ellipse is
$\frac{x\cos \theta }{a}+y\sin \theta =1$
Let this line cut the hyperbola at $A$ and $B$ and $O$ be the origin, then homogenizing of the hyperbola with the line
$a^2{x}^2-y^2={(\frac{x\cos \theta }{a}+y\sin \theta )}^2$
This is the equation of the pair of straight line $OA$ and $OB$
$\because \angle AOB=\frac{\pi }{2}$
$\therefore \text{coefficient of}{x}^2+\text{coefficient of}{y}^2=0\Rightarrow a^2-\frac{{\cos }^2\theta }{a^2}-1-{\sin }^2\theta =0\Rightarrow {\cos }^2\theta =\frac{a^2(2-a^2)}{(a^2-1)}\Rightarrow 0\le \frac{a^2(2-a^2)}{(a^2-1)}\le 1$
Let $b=a^2\Rightarrow b>0$
$\text{Now}0\le \frac{b(2-b)}{b-1}\le 1\Rightarrow \frac{(2-b)}{b-1}\ge 0b\in (1,2]...\left(1\right)\Rightarrow \frac{b(2-b)}{b-1}\le 1\frac{(b-b^2+1)}{b-1}\le 0b\in (0,1)\cup (\frac{\sqrt{5}+1}{2},\infty )...\left(2\right)\therefore a^2\in [\frac{\sqrt{5}+1}{2},2]$