Animated Solution for Mathematics - Conic Sections: 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

Welcome, fellow traveler of the coordinate plane. Today, we are not just solving a problem; we are learning how to read the language of curves.

Imagine you are standing on a vast, flat grid. You have two distinct shapes drawn on this grid: a perfect circle $C$ defined by $x^2+y^2=9$ and a graceful ellipse $E$ defined by:

Your mission is to determine the 'home' of two points, $P(1,2)$ and $Q(2,1)$, relative to these shapes. Are they trapped inside, or are they wandering outside?

Before we dive into the numbers, let us unlock the secret of the 'test function.' In coordinate geometry, any curve can be written as $S(x,y)=0$. This is our boundary.

But what happens when we plug a point $(x_1,y_1)$ into this expression? If $S(x_1,y_1)<0$, the point is inside the curve. If $S(x_1,y_1)>0$, the point is outside.

It is that simple, yet it is the most powerful tool in your arsenal. It turns a visual problem into a clear, binary decision.

Let us start with point $P(1,2)$. First, we test it against the circle $C$. We define our test function as $S_C=x^2+y^2-9$.

Plugging in our coordinates, we get $S_C(P)=(1{)}^2+(2{)}^2-9$. This simplifies to $1+4-9$, which is $-4$.

Because $-4<0$, we know with absolute certainty that $P$ is inside the circle.

Now, let us see how $P$ fares against the ellipse $E$. We define $S_E=\frac{x^2}{9}+\frac{y^2}{4}-1$.

Substituting $P(1,2)$, we find:

Since $\frac{1}{9}>0$, point $P$ is officially outside the ellipse. We have successfully mapped $P$!

Now, let us turn our attention to $Q(2,1)$. We repeat our process. For the circle, $S_C(Q)=(2{)}^2+(1{)}^2-9$, which is $4+1-9=-4$.

Again, a negative result! $Q$ is inside the circle.

But what about the ellipse? This is where the math gets interesting. We calculate $S_E(Q)=\frac{2^2}{9}+\frac{1^2}{4}-1$.

To solve this, we find a common denominator of $36$:

Look at that! The result is negative. This means $Q$ is inside the ellipse.

We have walked through the grid, tested our points, and decoded the signs. We found that $P$ is inside the circle but outside the ellipse, while $Q$ is inside both.

When we look at our options, we see that Option 4 perfectly matches our findings for point $P$.

Remember, in JEE Advanced, the beauty is not just in the final answer, but in the confidence you gain by mastering these fundamental techniques. You didn't just guess; you calculated, you verified, and you conquered.