Question

lf $P$ is a point on thc ellipse of eccentricity $e$ and $A,A^{{}^{\prime }}$ are the vertices and $S$, $S^{\prime }$ are the foci then $\Delta SPS{}^{\prime }:\Delta AP{A}^{{}^{\prime }}$ is

• A. $e^3:1$
• B. $e^2:1$
• C. $e:1$
• D. $2e:1$

Answer 1

The correct option is C $e:1$

As shown in the given figure, let the point $P$ on ellipse be $(a,b)$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$S$ and $S^{\prime }$ are the foci and $A$ and $A^{\prime }$ are the ends of major axis.

Given eccentricity $e=\sqrt{1-\frac{a^2}{b^2}}$

Now we know the coordinates of $A,A^{\prime },S$ and $S^{\prime }$.

$A:(a,0),A^{\prime }:(-a,0),S:(+ae,0),S^{\prime }:(-ae,0)$

Now we know all the 3 vertices of the triangles APA' and SPS'

$\Rightarrow area\left(\Delta AP{A}^{\prime }\right)=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ x& a& -a\\ y& 0& 0\end{array}\right|=ay$ .....(1)

and $area\left(\Delta SP{S}^{\prime }\right)=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ x& ae& -ae\\ y& 0& 0\end{array}\right|=aey$ ....(2)

So, the required ratio $\frac{area\left(\Delta SP{S}^{\prime }\right)}{area\left(\Delta AP{A}^{\prime }\right)}=\frac{aey}{ay}=e:1$