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Let E be the ellipe $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and $C$ be the circle $x^{2}+y^{2}=9$. Let $P=(1,2)$ and $Q=(2,1) .$ Which one of the following
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The correct answer is:
$P$ lies inside $C$ but outside $E$.
Given equation of ellipse $\mathrm{E}$ is $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
$\Rightarrow \frac{4 x^{2}+9 y^{2}}{36}=1 \Rightarrow 4 x^{2}+9 y^{2}=36$
$\Rightarrow 4 x^{2}+9 y^{2}-36=0$ $\ldots(1)$
and $C:$ Eqn of circle is $x^{2}+y^{2}=9$ Which can be rewritten as
$x^{2}+y^{2}-9=0$ $\ldots(2)$
For a point $P(1,2)$ we have
$4(1)^{2}+9(2)^{2}-36=40-36>0 \quad[$ from $(1)]$
and $1^{2}+2^{2}-9=5-9 < 0 \quad[$ from $(2)]$
Point $P$ lies outside of $E$ and inside of $C$.
$\Rightarrow \frac{4 x^{2}+9 y^{2}}{36}=1 \Rightarrow 4 x^{2}+9 y^{2}=36$
$\Rightarrow 4 x^{2}+9 y^{2}-36=0$ $\ldots(1)$
and $C:$ Eqn of circle is $x^{2}+y^{2}=9$ Which can be rewritten as
$x^{2}+y^{2}-9=0$ $\ldots(2)$
For a point $P(1,2)$ we have
$4(1)^{2}+9(2)^{2}-36=40-36>0 \quad[$ from $(1)]$
and $1^{2}+2^{2}-9=5-9 < 0 \quad[$ from $(2)]$
Point $P$ lies outside of $E$ and inside of $C$.
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