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Question Number 46838 by peter frank last updated on 01/Nov/18

	Answered by MrW3 last updated on 03/Nov/18

	$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $\frac{2 x}{a^{2}} + \frac{2 y}{b^{2}} y^{'} = 0 \Rightarrow \frac{x}{a^{2}} + \frac{y}{b^{2}} y^{'} = 0$ $e^{2} = \frac{a^{2} - b^{2}}{a^{2}}$ $e q n . o f c i r c l e :$ ${(x - c)}^{2} + y^{2} = r^{2} (s a y)$ $2 (x - c) + 2 {y y}^{'} = 0 \Rightarrow (x - c) + {y y}^{'} = 0$ $c o n t a c t p o i n t s : (e a, k) a n d (e a, - k)$ $\frac{e^{2} a^{2}}{a^{2}} + \frac{k^{2}}{b^{2}} = 1$ $e^{2} + \frac{k^{2}}{b^{2}} = 1$ $\Rightarrow e^{2} b^{2} + k^{2} = b^{2} . . . (i)$ ${(e a - c)}^{2} + k^{2} = r^{2} . . . (i i)$ $\frac{e a}{a^{2}} + \frac{k}{b^{2}} m = 0$ $\frac{{e b}^{2}}{a} + k m = 0 . . . (i i i)$ $e a - c + k m = 0 . . . (i v)$ $(i) - (i i) :$ $e^{2} b^{2} - {(e a - c)}^{2} = b^{2} - r^{2} . . . (v)$ $(i i i) - (i v) :$ $\frac{{e b}^{2}}{a} - e a + c = 0$ $\Rightarrow c = e (a - \frac{b^{2}}{a}) = \frac{e (a^{2} - b^{2})}{a} = e^{3} a$ $f r o m (v) :$ $r^{2} = b^{2} (1 - e^{2}) + {(e a - c)}^{2} = b^{2} (1 - e^{2}) + \frac{e^{2} b^{4}}{a^{2}} = b^{2} [1 - \frac{e^{2} (a^{2} - b^{2}}{a^{2}}] = b^{2} (1 - e^{4}) = a^{2} (1 - e^{2}) (1 - e^{4}) = a^{2} (1 - e^{2} - e^{4} + e^{6})$ $\Rightarrow e q n . o f c i r c l e :$ ${(x - e^{3} a)}^{2} + y^{2} = b^{2} (1 - e^{4})$ $y^{2} + x^{2} - 2 {a e}^{3} x + a^{2} e^{6} = a^{2} (1 - e^{2} - e^{4} + e^{6})$ $\Rightarrow y^{2} + x^{2} - 2 {a e}^{3} x = a^{2} (1 - e^{2} - e^{4})$
	Commented by peter frank last updated on 04/Nov/18

	$thank you {mrW}_{3}$
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