Question

The segment of the tangent at the point P to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, intercepted by the auxiliary circle subtends a right angle at the origin. If the eccentricity of the ellipse is smallest possible, then the point P can be

• A. $(0,ae)$
• B. $(a,0)$
• C. $(-a,0)$
• D. $(0,-b)$

Answer 1

The correct option is D $(0,-b)$

Equation of tangent at $P(acos\theta ,bsin\theta )$ is
$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
$\therefore$ Equation of the lines OA and OB is
$x^2+y^2-a^2{(\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b})}^2=0$
since the lines OA and OB are perpendicular
$\therefore$ $1-{\cos }^2\theta +1-\frac{a^2}{b^2}{\sin }^2\theta =0$
i.e ${\sin }^2\theta +1-\frac{a^2}{b^2}{\sin }^2\theta$
i.e $\frac{b^2}{a^2}=\frac{si{n}^2\theta }{1+{\sin }^2\theta }$
$\therefore e^2=1-\frac{b^2}{a^2}=1-\frac{{\sin }^2\theta }{1+{\sin }^2\theta }=\frac{1}{1+{\sin }^2\theta }$
$\therefore$ e is least if $\theta =\pm \frac{\pi }{2}$
$\therefore$ the point P is $(0,\pm b)$