Question

The set of values of ${\alpha }^2$, if there exists a tangent to the ellipse $\frac{x^2}{{\alpha }^2}+y^2=1$ such that the portion of the tangent intercepted by the hyperbola ${\alpha }^2{x}^2-y^2=1$ subtends a right angle at the centre of the curves, is

• A. $[\frac{\sqrt{5}+1}{2},2]$
• B. $[1,2]$
• C. $[\frac{\sqrt{5}-1}{2},1]$
• D. $[\frac{\sqrt{5}-1}{2},\frac{\sqrt{5}+1}{2}]$

Answer 1

The correct option is A $[\frac{\sqrt{5}+1}{2},2]$

Equation of any tangent to $\frac{x^2}{{\alpha }^2}+y^2=1$ in slope form is, $y=mx\pm \sqrt{{\alpha }^2{m}^2+1}$
$\Rightarrow {\left(\frac{y-mx}{\sqrt{{\alpha }^2{m}^2+1}}\right)}^2=1$
By homogenisation, we can write
$({\alpha }^2{x}^2-y^2)({\alpha }^2{m}^2+1)=y^2+m^2{x}^2-2mxy$
The portion of tangent subtends ${90}^{\circ }$ at the origin.
$\text{coefficient of}{x}^2+\text{coefficient of}{y}^2=0$
$\Rightarrow ({\alpha }^2{m}^2+1)({\alpha }^2-1)=1+m^2$
$\Rightarrow {\alpha }^2({\alpha }^2-1)m^2+{\alpha }^2-1=1+m^2$
$\Rightarrow \left[{\alpha }^2\right({\alpha }^2-1)-1]m^2=2-{\alpha }^2$
$\Rightarrow m^2=\frac{2-{\alpha }^2}{{\alpha }^4-{\alpha }^2-1}\ge 0$
$\Rightarrow \frac{{\alpha }^2-2}{{\alpha }^4-{\alpha }^2-1}\le 0$

Let ${\alpha }^2=t\ge 0$
Then $\frac{t-2}{t^2-t-1}\le 0$
$\Rightarrow t\in [\frac{1+\sqrt{5}}{2},2]$