Question

$p\left(\frac{\pi }{6}\right)$ is a point on the ellipse $\frac{x^2}{36}+\frac{y^2}{9}=1,S,S^1$ are foci of ellipse then $\left|Sp-S^{1}p\right|=$

• A. $6\sqrt{3}$
• B. 9
• C. $9\sqrt{3}$
• D. 27

Answer 1

The correct option is B 9

Given equation of ellipse is $\frac{x^2}{36}+\frac{y^2}{9}=1$

$\therefore a=6$ and $b=3$

Given $P\left(\frac{\pi }{6}\right)$ is the point on ellipse.

Thus, coordinates of point P are $P(6cos\frac{\pi }{6},3sin\frac{\pi }{6})$

$=P(6\frac{\sqrt{3}}{2},\frac{3}{2})$

$=P(3\sqrt{3},\frac{3}{2})$

Now, $e=\sqrt{1-\frac{b^2}{a^2}}$

$\therefore e=\sqrt{1-\frac{9}{36}}$

$\therefore e=\sqrt{\frac{27}{36}}$

$\therefore e=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}$

Thus, coordinates of foci are,

$S(ae,0)=S(6\times \frac{\sqrt{3}}{2},0)=S(3\sqrt{3},0)$

$S^{{}^{\prime }}(-ae,0)=S^{{}^{\prime }}(-3\sqrt{3},0)$

Thus, by distance formula,

$SP=\sqrt{{(3\sqrt{3}-3\sqrt{3})}^2+{(0-\frac{3}{2})}^2}$

$\therefore SP=\sqrt{{\left(0\right)}^2+{(-\frac{3}{2})}^2}=\frac{3}{2}$

Similarly, $S^{{}^{\prime }}P=\sqrt{{(-3\sqrt{3}-3\sqrt{3})}^2+{(0-\frac{3}{2})}^2}$

$\therefore S^{{}^{\prime }}P=\sqrt{{(-6\sqrt{3})}^2+{(-\frac{3}{2})}^2}$

$\therefore S^{{}^{\prime }}P=\sqrt{108+\frac{9}{4}}$

$\therefore S^{{}^{\prime }}P=\sqrt{\frac{441}{4}}=\frac{21}{2}$

$\therefore |SP-S^{{}^{\prime }}P|=|\frac{3}{2}-\frac{21}{2}|$

$\therefore |SP-S^{{}^{\prime }}P|=\left|\frac{-18}{2}\right|=9$