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