Animated Solution for Mathematics - Conic Sections: Let $S=\{(x,y)\in \mathbb{N}\times \mathbb{N}:9(x-3{)}^2+16(y-4{)}^2\le 144\}$ and $T=\{(x,y)\in \mathbb{R}\times \mathbb{R}:(x-7{)}^2+(y-4{)}^2\le 36\}$. The $n(S\cap T)$ is equal to

The problem asks us to find the number of integer points $(x,y)$ that satisfy both the elliptical constraint $S$ and the circular constraint $T$. The sets are defined as:

In JEE Advanced, complexity is often a mask for elegance. Our goal is to peel back that mask by simplifying the coordinate system.

Solving this in the original $(x,y)$ coordinates is cumbersome because the centers are offset. Let us define new coordinates $X=x-3$ and $Y=y-4$.

The ellipse equation transforms into:

Since $x,y\in \mathbb{N}$, we have $x\ge 1$ and $y\ge 1$. This implies the constraints $X\ge -2$ and $Y\ge -3$.

The ellipse is the more restrictive boundary, limiting $Y$ to the interval $[-3,3]$. We will test each integer value of $Y$ within this range to find valid $X$ values that satisfy both the ellipse and the circle $(X-4{)}^2+Y^2\le 36$.

For $Y=0$: The ellipse gives $X^2\le 16$, so $X\in [-4,4]$. The circle gives $(X-4{)}^2\le 36$, so $X\in [-2,10]$. Considering $X\ge -2$, the valid integers are $X\in \{-2,-1,0,1,2,3,4\}$. This yields 7 points.

For $Y=\pm 1$: The ellipse gives $X^2\le 16(1-1/9)=128/9\approx 14.22$, so $X\in [-3.77,3.77]$. The circle gives $(X-4{)}^2\le 35$, so $X\in [4-\sqrt{35},4+\sqrt{35}]\approx [-1.91,9.91]$. The intersection with $X\ge -2$ gives $X\in \{-1,0,1,2,3\}$. This yields 5 points for each $Y$, totaling 10 points.

For $Y=\pm 2$: The ellipse gives $X^2\le 16(1-4/9)=80/9\approx 8.88$, so $X\in [-2.98,2.98]$. The circle gives $(X-4{)}^2\le 32$, so $X\in [4-\sqrt{32},4+\sqrt{32}]\approx [-1.65,9.65]$. The intersection with $X\ge -2$ gives $X\in \{-1,0,1,2\}$. This yields 4 points for each $Y$, totaling 8 points.

For $Y=\pm 3$: The ellipse gives $X^2\le 16(1-9/9)=0$, so $X=0$. Checking the circle: $(0-4{)}^2+3^2=16+9=25\le 36$. This is valid. This yields 1 point for each $Y$, totaling 2 points.

Summing the points from all rows:

The total number of integer points satisfying the given conditions is 27.