Let E be the ellipse frac{x^2}{9} + frac{y^2} | vedclass

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(D) The equation of the ellipse is $E: \frac{x^2}{9} + \frac{y^2}{4} = 1$ and the circle is $C: x^2 + y^2 = 9$.For point $P(1, 2)$:Check for circle $C

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$P$ lies inside $C$ but outside $E$

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(D) The equation of the ellipse is $E: \frac{x^2}{9} + \frac{y^2}{4} = 1$ and the circle is $C: x^2 + y^2 = 9$.For point $P(1, 2)$:Check for circle $C$: $1^2 + 2^2 = 1 + 4 = 5 Check for ellipse $E$: $\frac{1^2}{9} + \frac{2^2}{4} = \frac{1}{9} + 1 = \frac{10}{9} > 1$,so $P$ is outside $E$.For point $Q(2, 1)$:Check for circle $C$: $2^2 + 1^2 = 4 + 1 = 5 Check for ellipse $E$: $\frac{2^2}{9} + \frac{1^2}{4} = \frac{4}{9} + \frac{1}{4} = \frac{16+9}{36} = \frac{25}{36} Thus,$P$ lies inside $C$ but outside $E$.

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:

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