Conic Sections Question 272
Question: 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 $ PS_1S_2 $ lies on
Options:
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
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Answer:
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 $ (acos\theta ,b\sin \theta ). $ The coordinates of foci are $ S_1(ae,0) $ and $ S_2(-ae,0). $
Hence, $ S_1P=a(1-ecos\theta ) $ $ S_2P=a(1+ecos\theta ) $ and $ S_1S_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}}{4a^{2}}+\frac{k^{2}}{{b^{^{2}}}}={{( \frac{e}{1+e} )}^{2}} $
$ \therefore $ The locus of the point $ (h,k) $ is $ \frac{x^{2}}{4a^{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}}{4a^{2}}}=\sqrt{\frac{3+e^{2}}{4}} $
BETA
