Question

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

• A. $\left[\begin{array}{c}\frac{\sqrt{5}+1}{2},2\end{array}\right]$
• B. $(1,2]$
• C. $[\frac{\sqrt{5}-1}{2},1)$
• D. $[\frac{\sqrt{5}-1}{2},1)\cup (1,\frac{\sqrt{5}+1}{2}]$

Answer 1

The correct option is A $\left[\begin{array}{c}\frac{\sqrt{5}+1}{2},2\end{array}\right]$

Equation of tangent at point $P(a\cos \theta ,\sin \theta )$ is
$\frac{x}{a}\cos \theta +\frac{y}{1}\sin \theta =1\cdots \left(1\right)$
Let, it cut the hyperbola at points $P$ and $Q$
Homogenizing the hyperbola $a^2{x}^2-y^2=1$ with the help of the above the equation, we get
$a^2{x}^2-y^2={(\frac{x}{a}\cos \theta +y\sin \theta )}^2$
This is a pair of straight lines $OP$ and $OQ$
$\because OP\perp OQ$
$\Rightarrow \text{Coefficient of}{x}^2+\text{coefficient of}{y}^2=0$
$\Rightarrow a^2-\frac{{\cos }^2\theta }{a^2}-1-{\sin }^2\theta =0$
$\Rightarrow a^2-\frac{{\cos }^2\theta }{a^2}-1-1+{\cos }^2\theta =0$
$\Rightarrow {\cos }^2\theta =\frac{a^2(2-a^2)}{a^2-1}$
$\text{Now,}$ $0\le {\cos }^2\theta \le 1$
$\Rightarrow 0\le \frac{a^2(2-a^2)}{a^2-1}\le 1$
Let $t=a^2,t>0$
Case $\left(i\right)$ $0\le \frac{t(2-t)}{(t-1)}\Rightarrow \frac{t(t-2)}{(t-1)}\le 0$
$\Rightarrow t\in (1,2]\cdots \left(1\right)(\because t>0)$
Case $\left(ii\right)\frac{t(2-t)}{(t-1)}\le 1$
$\Rightarrow \frac{-t^2+t+1}{t-1}\le 0$
$\Rightarrow \frac{(t-\frac{1+\sqrt{5}}{2})(t-\frac{1-\sqrt{5}}{2})}{(t-1)}\ge 0$
$\Rightarrow t\in [0,1)\cup [\frac{1+\sqrt{5}}{2},\infty )\cdots \left(2\right)(\because t>0)$
Hence required solution is
$\therefore a^2\in \left[\begin{array}{c}\frac{\sqrt{5}+1}{2},2\end{array}\right]$