An ellipse is drawn with major and minor axes | vedclass

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Question

$2a = 10 \Rightarrow a = 5 ; 2b = 8 \Rightarrow b = 4$ 
$e^2 = 1 - \frac{{16}}{{25}}\,$=$\frac{9}{{25}}\, \Rightarrow e = \frac{3}{5}\,$&nb

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$2$

Vedclass · Answer

$2a = 10 \Rightarrow a = 5 ; 2b = 8 \Rightarrow b = 4$ 
$e^2 = 1 - \frac{{16}}{{25}}\,$=$\frac{9}{{25}}\, \Rightarrow e = \frac{3}{5}\,$ 
Focus = $(3, 0)$ 
Let the circle touches the ellipse at $ P $ and $Q$. Consider a tangent (to both circle and ellipse) at $P$. Let $F$(one focus) be the centre of the circle and other focus be $G.$  A ray from $F$  to $ P$ must retrace its path (normal to the circle). But the reflection propety the ray $FP$  must be reflected along $PG$. This is possible only if $ P, F $ and $G $ are collinear. Thus $P$ must be the end of the major axis.Hence $ r = a - ae = 5 - 3 = 2$ 
  alternately normal to an ellipse at $P$  must pass through the centre $(3, 0)$  of the circle 
$\frac{{ax}}{{\cos 	heta }}\,\, - \,\,\frac{{by}}{{\sin 	heta }}\,\, = \,\,{a^2} - {b^2} $ $\Rightarrow \frac{{5x}}{{\cos 	heta }}\,\, - \,\,\frac{{4y}}{{\sin 	heta }}\,\, = \,\,9$ $\left( {	heta \, 
e \,0\,\,or\,\frac{\pi }{2}\,} ight)\,$
$\frac{{15}}{{\cos 	heta }}\,\, - \,\,0\, = 9\, \Rightarrow \cos 	heta \, = \,\frac{{15}}{9}\,$  $\Rightarrow$ which is not possible $\Rightarrow 	heta\ = 0 or \pi /2$ 
but $	heta \pi /2 \Rightarrow 	heta = 0$
Hence $P (5, 0)$ i.e. end of major axis

An ellipse is drawn with major and minor axes of lengths $10 $ and $8$ respectively. Using one focus as centre, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse. The radius of the circle is

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