Question
If a variable point $$P$$ on an ellipse of eccentricity $$e$$ is joined to the foci $${S_1}$$ and $${S_2}$$ then the incentre 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
Answer :
Another ellipse of eccentricity $$\sqrt {\frac{{3 + {e^2}}}{4}} $$
Solution :
Let the ellipse be $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1......\left( 1 \right)$$
Then $${e^2} = 1 - \frac{{{b^2}}}{{{a^2}}}......\left( 2 \right)$$
Let a point $$P$$ on $$\left( 1 \right)$$ be $$\left( {a\,\cos \,\theta ,\,b\,\sin \,\theta } \right)$$
The coordinates of foci are $${S_1}\left( {ae,\,0} \right)$$ and $${S_2}\left( { - ae,\,0} \right)$$
Hence, $${S_1}P = a\left( {1 - e\,\cos \,\theta } \right)$$
$${S_2}P = a\left( {1 + e\,\cos \,\theta } \right)$$ and $${S_1}{S_2} = 2ae$$
If $$\left( {h,\,k} \right)$$ be the coordinates of in centre then
$$\eqalign{ & h = \frac{{2ae \times a\,\cos \,\theta + a\left( {1 - e\,\cos \,\theta } \right) \times - ae + a\left( {1 + e\,\cos \,\theta } \right) \times ae}}{{2ae + a\left( {1 - e\,\cos \,\theta } \right) + a\left( {1 + e\,\cos \,\theta } \right)}} \cr & = \frac{{2ae\,\cos \,\theta }}{{1 + e}}......\left( 3 \right) \cr & k = \frac{{be\,\sin \,\theta }}{{1 + e}}......\left( 4 \right) \cr} $$
Squaring and adding $$\left( 3 \right)$$ & $$\left( 4 \right)$$ 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 $$\left( {h,\,k} \right)$$ is
$$\frac{{{x^2}}}{{4{a^2}{\lambda ^2}}} + \frac{{{y^2}}}{{{b^2}{\lambda ^2}}} = 1,{\text{ where }}\lambda = \frac{e}{{1 + e}}$$
Which is another ellipse with eccentricity $$ = \sqrt {1 - \frac{{{b^2}}}{{4{a^2}}}} = \sqrt {\frac{{3 + {e^2}}}{4}} $$
Let the ellipse be $$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1......\left( 1 \right)$$
Then $${e^2} = 1 - \frac{{{b^2}}}{{{a^2}}}......\left( 2 \right)$$
Let a point $$P$$ on $$\left( 1 \right)$$ be $$\left( {a\,\cos \,\theta ,\,b\,\sin \,\theta } \right)$$
The coordinates of foci are $${S_1}\left( {ae,\,0} \right)$$ and $${S_2}\left( { - ae,\,0} \right)$$
Hence, $${S_1}P = a\left( {1 - e\,\cos \,\theta } \right)$$
$${S_2}P = a\left( {1 + e\,\cos \,\theta } \right)$$ and $${S_1}{S_2} = 2ae$$
If $$\left( {h,\,k} \right)$$ be the coordinates of in centre then
$$\eqalign{ & h = \frac{{2ae \times a\,\cos \,\theta + a\left( {1 - e\,\cos \,\theta } \right) \times - ae + a\left( {1 + e\,\cos \,\theta } \right) \times ae}}{{2ae + a\left( {1 - e\,\cos \,\theta } \right) + a\left( {1 + e\,\cos \,\theta } \right)}} \cr & = \frac{{2ae\,\cos \,\theta }}{{1 + e}}......\left( 3 \right) \cr & k = \frac{{be\,\sin \,\theta }}{{1 + e}}......\left( 4 \right) \cr} $$
Squaring and adding $$\left( 3 \right)$$ & $$\left( 4 \right)$$ 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 $$\left( {h,\,k} \right)$$ is
$$\frac{{{x^2}}}{{4{a^2}{\lambda ^2}}} + \frac{{{y^2}}}{{{b^2}{\lambda ^2}}} = 1,{\text{ where }}\lambda = \frac{e}{{1 + e}}$$
Which is another ellipse with eccentricity $$ = \sqrt {1 - \frac{{{b^2}}}{{4{a^2}}}} = \sqrt {\frac{{3 + {e^2}}}{4}} $$