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Q.
Let $E$ be the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and $C$ be the circle $x^2+y^2=9$. Let $P$ and $Q$ be the points $(1,2)$ and $(2,1)$, respectively. Then
Conic Sections
Solution:
Since $1^2+2^2-9<0$ and $2^2+1^2-9<0$, both $P$ and $Q$ lie inside $C$.
Also $\frac{1^2}{9}+\frac{2^2}{4}-1>0$ and $\frac{2^2}{9}+\frac{1^2}{4}-1<0$, P lies outside $E$ and $Q$ lies inside $E$.
Thus, $P$ lies inside $C$ but outside E.