Animated Solution for Mathematics - Conic Sections: If $x^2+9y^2-4x+3=0,x,y\in \mathbb{R}$, then $x$ and $y$ respectively lie in the intervals:

Unmasking the Ellipse

Welcome, my dear student. Today, we are not just solving an equation; we are embarking on a journey of discovery.

We are given the expression $x^2+9y^2-4x+3=0$ and told that $x$ and $y$ are real numbers. At first glance, this looks like a jumbled mess of variables and constants. But in the world of JEE Advanced, we don't see messes; we see potential.

We see a story waiting to be told.

The Art of Algebraic Surgery

Our first instinct must be to bring order to chaos. We have $x$ terms scattered about, and $y$ terms sitting in isolation. Let us group them.

We take the $x^2$ and the $-4x$ and place them in a bracket: $(x^2-4x)$. The $9y^2$ and the constant $3$ can wait their turn. Now, we perform the most elegant maneuver in algebra: completing the square.

Look at the coefficient of $x$, which is $-4$. We take half of it, which is $-2$, and square it to get $4$. We add and subtract this $4$ inside our equation.

Why? Because we want to create a perfect square trinomial. The equation becomes $(x^2-4x+4)-4+9y^2+3=0$.

Notice how the first three terms collapse into $(x-2{)}^2$. We are left with $(x-2{)}^2+9y^2-1=0$. Suddenly, the chaos has vanished, and a beautiful, recognizable structure emerges.

The Geometric Revelation

Let us move the constant to the right side: $(x-2{)}^2+9y^2=1$. Does this look familiar? It should! This is the standard form of an ellipse, shifted from the origin.

To see it clearly, we can write it as:

We have a center at $(2,0)$, a semi-major axis of $a=1$, and a semi-minor axis of $b=\frac{1}{3}$. Imagine this ellipse on a coordinate plane.

It is a flattened circle, stretched horizontally, centered at $x=2$ on the x-axis. It is a bounded, finite shape. This is why we can talk about intervals for $x$ and $y$—because the ellipse does not go on forever; it is trapped within a box.

The Bounding Box

Now, we come to the heart of the problem. We need to find the range of $x$ and $y$. Since $x$ and $y$ are real, we know that any squared term must be non-negative.

Let us use this to our advantage. From our equation $(x-2{)}^2+9y^2=1$, we can isolate $9y^2$:

Since $y$ is a real number, $y^2\ge 0$, which implies $9y^2\ge 0$. Therefore, the right side must also be non-negative:

This inequality is our key. It tells us that $(x-2{)}^2\le 1$. Taking the square root, we get $-1\le x-2\le 1$.

Adding $2$ to all parts, we find that $1\le x\le 3$. This is the horizontal boundary of our ellipse!

We repeat this logic for $y$. Isolate the $x$ term: $(x-2{)}^2=1-9y^2$. Since $(x-2{)}^2\ge 0$, it follows that $1-9y^2\ge 0$.

Rearranging this, we get $9y^2\le 1$, or $y^2\le \frac{1}{9}$. Taking the square root, we find $-\frac{1}{3}\le y\le \frac{1}{3}$.

The Beauty of Constraints

We have arrived at our destination. The values of $x$ are confined to the interval $[1,3]$, and the values of $y$ are confined to $[-\frac{1}{3},\frac{1}{3}]$.

We didn't just solve an equation; we defined the boundaries of a geometric object. Remember, my student, that in physics and mathematics, constraints are not limitations—they are the definitions of the system.