Question

The tangent and normal at a point $P$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the minor axis is $A$ and $B.$ $S$ and $S^{\prime }$ are the foci of the ellipse,then:

• A. $\angle APB=\frac{\pi }{2}$
• B. $\angle ASB=\frac{\pi }{2}$
• C. $\angle A{S}^{\prime }B=\frac{\pi }{2}$
• D. None of these

Answer 1

Answer: (A), (B), (C)

$\angle APB=\frac{\pi }{2}$
$\angle A{S}^{\prime }B=\frac{\pi }{2}$
$\angle ASB=\frac{\pi }{2}$

Let the point $P$ be $(a\cos \theta ,b\sin \theta )$

Tangent $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$ meets minor axis at $(0,\frac{b}{\sin \theta })$

$\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2{b}^2$ meets minor axis at $(0,\frac{-\sin \theta }{b}(a^2-b^2\left)\right)$

$\angle APB=90°\left(\frac{\pi }{2}\right)$ because tangent of ellipse is normal to the ellipse

$m_{AS}{m}_{BS}=\frac{ae-0}{\frac{-b}{\sin \theta }}\times \frac{ae-0}{\frac{\sin \theta }{b}(a^2-b^2)}$

$m_{AS}{m}_{BS}=\frac{a^2{e}^2}{-(a^2-b^2)}=-1$

So, $AS$ is perpendicular to $SB$.

Thus $\angle ASB=90°$

Similarly $\angle A{S}^{\prime }B=90°$.