Question

The tangent at any point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the auxiliary circle at two points which subtend a right angle at the centre. When the eccentricity of the ellipse is minimum then

• A. the $y-$ intercept made by the tangent is $y=\pm b\sqrt{2}$ when $a>b$
• B. the $x-$ intercept made by the tangent is $x=\pm a\sqrt{2}$ when $a<b$
• C. minimum value of eccentricity is $\frac{1}{\sqrt{2}}$
• D. When $a>b$, then the area enclosed between the tangents and the ellipse is $(4-\pi )ab$

Answer 1

Answer: (C), (D)

The correct options are
C minimum value of eccentricity is $\frac{1}{\sqrt{2}}$
D When $a>b$, then the area enclosed between the tangents and the ellipse is $(4-\pi )ab$

When $a>b$
Equation of the Auxiliary circle is,
$x^2+y^2=a^2\cdots \left(1\right)$
Equation of tangent at any point $P(a\cos \alpha ,b\sin \alpha )$ is,
$\frac{x\cos \alpha }{a}+\frac{y\sin \alpha }{b}=1\cdots \left(2\right)$


Making equation $\left(2\right)$ homogeneous using equation $\left(1\right)$,
$x^2+y^2-a^2{(\frac{x\cos \alpha }{a}+\frac{y\sin \alpha }{b})}^2=0(1-{\cos }^2\alpha )x^2+(1-\frac{a^2}{b^2}{\sin }^2\alpha )y^2-\left(\frac{2a}{b}\sin \alpha \cos \alpha \right)xy=0\cdots \left(3\right)$
Equation $\left(3\right)$ represents the combine equation of the line $OL\text{and}OM$,
Since, $\angle LOM=\frac{\pi }{2}$
$\text{Coefficient of}{x}^2+\text{Coefficient of}{y}^2=0$
$\Rightarrow (1-{\cos }^2\alpha )+(1-\frac{a^2}{b^2}{\sin }^2\alpha )=0\Rightarrow {\sin }^2\alpha (\frac{a^2}{b^2}-1)=1\Rightarrow {\sin }^2\alpha (\frac{1}{1-e^2}-1)=1[\because b^2=a^2(1-e^2\left)\right]\Rightarrow e=\frac{1}{\sqrt{1+{\sin }^2\alpha }}$

When $b>a$
Equation of the Auxiliary circle is,
$x^2+y^2=b^2\cdots \left(1\right)$
Equation of tangent any point $P(a\cos \alpha ,b\sin \alpha )$
$\frac{x\cos \alpha }{a}+\frac{y\sin \alpha }{b}=1\cdots \left(2\right)$

Making equation $\left(2\right)$ homogeneous using equation $\left(1\right)$,
$x^2+y^2-b^2{(\frac{x\cos \alpha }{a}+\frac{y\sin \alpha }{b})}^2=0$

$\text{Coefficient of}{x}^2+\text{Coefficient of}{y}^2=0$
$\Rightarrow 1-\frac{b^2}{a^2}{\cos }^2\alpha +{\cos }^2\alpha =0\Rightarrow {\cos }^2\alpha (\frac{b^2}{a^2}-1)=1\Rightarrow {\cos }^2\alpha (\frac{1}{1-e^2}-1)=1[\because a^2=b^2(1-e^2\left)\right]\Rightarrow e=\frac{1}{\sqrt{1+{\cos }^2\alpha }}$

Therefore,
$a>b\Rightarrow e=\frac{1}{\sqrt{1+{\sin }^2\alpha }}b>a\Rightarrow e=\frac{1}{\sqrt{1+{\cos }^2\alpha }}$

Minimum value of eccentricity is
$e=\frac{1}{\sqrt{2}}$
When $a>b$
$\alpha =\frac{\pi }{2},\frac{3\pi }{2}$
Tangents $y=\pm b$
When $b>a$
$\alpha =0,\pi$
Tangents $x=\pm a$

Area bounded when $a>b$,
$A=2a\times 2b-\pi ab=(4-\pi )ab$