question_answer

If a variable point P on an ellipse of eccentricity e lines joining the foci \[{{S}_{1}}\] and \[{{S}_{2}}\] then the in centre of the triangle \[P{{S}_{1}}{{S}_{2}}\] lies on

A) The major axis of the ellipse

B) The circle with radius e

C) Another ellipse of eccentricity \[\sqrt{\frac{3+{{e}^{2}}}{4}}\]

D) None of these

Correct Answer: C

Solution :

[c]

Let the ellipse be \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]             ?. (1)

Then \[{{e}^{2}}=1-\frac{{{b}^{2}}}{{{a}^{2}}}\]                          ?. (2)

Let a point P on (1) be \[(a\,\,cos\,\theta ,b\,\,\sin \,\theta ).\]

The coordinates of foci are \[{{S}_{1}}(ae,0)\] and \[{{S}_{2}}(-ae,0).\]

Hence, \[{{S}_{1}}P=a(1-e\,\,cos\theta )\]

\[{{S}_{2}}P=a(1+e\,\,cos\theta )\] and \[{{S}_{1}}{{S}_{2}}=2ae\]

If \[(h,k)\] be the coordinates of in centre then

\[h=\frac{2ae\times a\cos \theta +a(1-ecos\theta )\times -ae+a(1+ecos\theta )\times ae}{2ae+a(1-ecos\theta )+a(1+ecos\theta )}\]\[=\frac{2ae\cos \theta }{1+e}\]                                           ? (3)

\[k=\frac{be\sin \theta }{1+e}\]                                       ?. (4)

Squaring and adding (3) & (4) we have.

\[\frac{{{h}^{2}}}{4{{a}^{2}}}+\frac{{{k}^{2}}}{{{b}^{^{2}}}}={{\left( \frac{e}{1+e} \right)}^{2}}\]

\[\therefore \] The locus of the point \[(h,k)\] is

\[\frac{{{x}^{2}}}{4{{a}^{2}}{{\lambda }^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}{{\lambda }^{2}}}=1,\] where \[\lambda =\frac{e}{1+e}\]

Which is another ellipse with eccentricity

\[=\sqrt{1-\frac{{{b}^{2}}}{4{{a}^{2}}}}=\sqrt{\frac{3+{{e}^{2}}}{4}}\]