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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
$Q$ lies inside $C$ but outside $E$
$Q$ lies outside both $C$ and $E$
$P$ lies inside both $C$ and $E$
$P$ lies inside $C$ but outside $E$
Solution
(d) The given ellipse is $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$.
The value of the expression $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} – 1$ is positive for $x = 1,\,y = 2$ and negative for $x = 2,\,y = 1$.
Therefore $P$ lies outside $E$ and $Q$ lies inside $E$.
The value of the expression ${x^2} + {y^2} – 9$ is negative for both the points $P$ and $Q$.
Therefore $P$ and $Q$ both lie inside $C$.
Hence $P$ lies inside $ C$ but outside $E.$
