Question

If the tangent at $\theta$ on the ellipse $x^2/a^2+y^2/b^2=1$ meets the auxiliary circle at two point which subtended a right angle at
the centre, then $e^2(2-co{s}^2\theta )=$

• A. 1
• B. 2
• C. -1
• D. 0

Answer 1

The correct option is A 1

We know that distance of any point on the ellipse from the foci of ellipse is given by,

$d=\frac{a^2\pm cx}{a}$

Refer the figure. Let coordinates of point P are $P(acos\theta ,bsin\theta )$

$\therefore A{F}_2=\frac{a^2+c\left(acos\theta \right)}{a}$

$\therefore A{F}_2=\frac{a^2+accos\theta }{a}$

$\therefore A{F}_2=a+c.cos\theta$

We know that, $e=\frac{c}{a}$

$\therefore c=ae$

$\therefore A{F}_2=a+aecos\theta$

Similarly, $A{F}_1=a-aecos\theta$

By distance formula, $F_1{F}_2=2c$

Now, ${A{F}_1}^2+{A{F}_2}^2={F_1{F}_2}^2$

$\therefore {(a-aecos\theta )}^2+{(a+aecos\theta )}^2={\left(2c\right)}^2$

$\therefore a^2-2a^{2}ecos\theta +a^2{e}^2{cos}^2\theta +a^2+2a^{2}ecos\theta +a^2{e}^2{cos}^2\theta =4c^2$

$\therefore 2a^2+2a^2{e}^2{cos}^2\theta =4c^2$

$\therefore 2a^2(1+e^2{cos}^2\theta )=4c^2$

We know that for an ellipse, $c^2=a^2-b^2$

$\therefore a^2(1+e^2{cos}^2\theta )=2(a^2-b^2)$

$\therefore a^2(1+e^2{cos}^2\theta )=2a^2-2b^2$

$\therefore a^2+a^2{e}^2{cos}^2\theta =2a^2-2b^2$

$\therefore a^2-a^2{e}^2{cos}^2\theta =2b^2$

$\therefore a^2(1-e^2{cos}^2\theta )=2b^2$

$\therefore 1-e^2{cos}^2\theta =\frac{2b^2}{a^2}$

$\therefore e^2{cos}^2\theta =1-\frac{2b^2}{a^2}$ (1)

Now, $e=\frac{c}{a}$

$\therefore e^2=\frac{c^2}{a^2}$

$\therefore e^2=\frac{a^2-b^2}{a^2}$

$\therefore e^2=1-\frac{b^2}{a^2}$

$\therefore \frac{b^2}{a^2}=1-e^2$

From equation (1), we get,

$e^2{cos}^2\theta =1-2(1-e^2)$

$\therefore e^2{cos}^2\theta =1-2+2e^2$

$\therefore e^2{cos}^2\theta =-1+2e^2$

$\therefore 2e^2-e^2{cos}^2\theta =1$

$\therefore e^2(2-{cos}^2\theta )=1$

Thus, answer is option (A)