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  • 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

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

Option: 1

\sqrt{3}


Option: 2

2


Option: 3

2 \sqrt{2}


Option: 4

\sqrt{5}


Answers (1)

best_answer

\mathrm{2 a=10 \Rightarrow a=5 ; 2 b=8 \Rightarrow b=4}

\mathrm{\mathrm{e}^2=1-\frac{16}{25}=\frac{9}{25} \Rightarrow \mathrm{e}=\frac{3}{5}}

Let the circle touches the ellipse at $P$ and Q. Consider a tangent (to both circle and ellipse) at P. Let \mathrm{\mathrm{F}} (one focus) be the centre of the circle and other focus be G. A ray from \mathrm{\mathrm{\mathrm{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 \mathrm{\mathrm{P}, \mathrm{F}} and \mathrm{\mathrm{G}} are collinear. Thus \mathrm{\mathrm{P}} must be the end of the major axis.

Hence \mathrm{ r=a-a e=5-3=2}

alternately normal to an ellipse at P must pass through the centre (3,0) of the circle

\mathrm{ \frac{a x}{\cos \theta}-\frac{\text { by }}{\sin \theta}=a^2-b^2 \Rightarrow \frac{5 x}{\cos \theta}-\frac{4 y}{\sin \theta}=9 \quad\left(\theta \neq 0 \text { or } \frac{\pi}{2}\right)}

\mathrm{\frac{15}{\cos \theta}-0=9 \Rightarrow \cos \theta=\frac{15}{9} \Rightarrow \text { which is not possible } \Rightarrow \theta=0 \text { or } \pi / 2}

\mathrm{\text { but } \theta \neq \pi / 2 \Rightarrow \theta=0}

Hence \mathrm{ P \equiv(5,0)}  i.e. end of major axis

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Ritika Harsh

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