Question

The tangent at any point $P(a\cos \theta ,b\sin \theta )$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the auxiliary circle at two point which subtend a right angle at the centre, then eccentricity is

• A. $\frac{1}{\sqrt{1+{\sin }^2\theta }}$
• B. $\frac{1}{\sqrt{2-{\cos }^2\theta }}$
• C. $\frac{1}{\sqrt{1+\tan 2\theta }}$
• D. $noneofthese$

Answer 1

The correct option is A $\frac{1}{\sqrt{1+{\sin }^2\theta }}$

consider the given equation of an ellipse

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

at any point $p(a\cos \theta ,b\sin \theta )$

Here, $\angle ROQ={90}^o$ (right angle)

and $\angle F_{1}P{F}_2={90}^o$ (right angle)

then,

$F_{1}P\parallel OQand{F}_{2}P\parallel OR$

now, we know that,

${\left|P{F}_1\right|}^2+{\left|P{F}_2\right|}^2=2a$ (by ellipse)

but

$\left|P{F}_1\right|=a+e{x}_1$

$\left|P{F}_2\right|=a-e{x}_2$

then,

$(a+e{x}_1{)}^2+(a+e{x}_2{)}^2=2a$

so,

$(a+e{x}_1{)}^2+(a+e{x}_2)=2a=\left|F_{!}{F}_2\right|$

$\Rightarrow (a+e{x}_1{)}^2+(a+e{x}_2)=4c^2$

so,

$a^2+e^2{x}_1^2=2c^2$

put $x_1=a\cos \theta$ and we get,

$\Rightarrow a^2+e^2(a\cos \theta {)}^2=2c^2$

$\Rightarrow a^2+a^2{e}^2{\cos }^2\theta =2c^2$

$\Rightarrow a^2[1+e^2{\cos }^2\theta ]=2c^2$

$\Rightarrow a^2[1+e^2(1-{\sin }^2\theta \left)\right]=2c^2$

and solve that.

$e^2(1++{\sin }^2\theta )=y$ (by ellipse)

$e^2=\frac{1}{1+{\sin }^2\theta }$

$e=\frac{1}{\sqrt{1+{\sin }^2\theta }}$

Hence, this is the answer.