Key Concept: Transforming General Quadratic Equations into Standard Conic Section Forms using Completing the Square

This problem requires us to find the permissible range of real values for $x$ and $y$ that satisfy a given quadratic equation. The most effective strategy for such problems is to transform the equation into the standard form of a conic section, if possible. In this case, we will find that the equation represents an ellipse. Once in standard form, the geometric properties of the ellipse directly reveal the maximum and minimum values (the range) for $x$ and $y$. The technique of completing the square is the fundamental algebraic tool for achieving this transformation.

The standard form of an ellipse centered at $(h, k)$ is given by: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ Here, $a$ and $b$ represent the lengths of the semi-major and semi-minor axes (or semi-axes along the x and y directions, respectively). For any point $(x, y)$ that lies on the boundary or within this ellipse, the following conditions must hold true:

• The $x$-coordinates range from $h-a$ to $h+a$, i.e., $x \in [h-a, h+a]$.
• The $y$-coordinates range from $k-b$ to $k+b$, i.e., $y \in [k-b, k+b]$. These ranges are derived from the fact that since both terms $\frac{(x-h)^2}{a^2}$ and $\frac{(y-k)^2}{b^2}$ are non-negative and their sum is 1, each individual term must be less than or equal to 1. For example, $\frac{(x-h)^2}{a^2} \le 1 \implies (x-h)^2 \le a^2 \implies |x-h| \le a$, which then leads to $h-a \le x \le h+a$.

---

Step-by-Step Solution:

1. Rearrange and Group Terms We begin with the given equation: $x^2 + 9y^2 - 4x + 3 = 0$ To prepare for completing the square, we group the terms involving $x$ together and the terms involving $y$ together, moving the constant term to the side: $(x^2 - 4x) + 9y^2 + 3 = 0$ Notice that the $y$-term, $9y^2$, is already a perfect square in terms of $y^2$ and lacks a linear $y$-term (like $cy$). This simplifies the process for $y$.

2. Complete the Square for x-terms Our goal is to transform the expression $(x^2 - 4x)$ into a perfect square of the form $(x-h)^2$. For a general quadratic expression $X^2 + BX$, we complete the square by adding $(B/2)^2$ to make it $(X + B/2)^2$. For $x^2 - 4x$, the coefficient of $x$ is $B = -4$. So, we need to add $(-4/2)^2 = (-2)^2 = 4$. To maintain the equality of the original equation, if we add a value to one side, we must also subtract it from the same side (or add it to the other side). $(x^2 - 4x + 4) - 4 + 9y^2 + 3 = 0$ Now, the terms inside the parenthesis form a perfect square: $(x^2 - 4x + 4) = (x - 2)^2$. Substitute this back into the equation: $(x - 2)^2 - 4 + 9y^2 + 3 = 0$ Combine the constant terms: $-4 + 3 = -1$. $(x - 2)^2 + 9y^2 - 1 = 0$

3. Transform to the Standard Form of an Ellipse To match the standard form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, we first move the constant term to the right side of the equation: $(x - 2)^2 + 9y^2 = 1$ Now, we need to express each term as a square divided by a constant squared.

• For the $x$-term, $(x-2)^2$ can be written as $\frac{(x-2)^2}{1}$. This implies that $a^2 = 1$, so $a = \sqrt{1} = 1$.
• For the $y$-term, we have $9y^2$. To get $y^2$ in the numerator with a denominator representing $b^2$, we can rewrite $9y^2$ as $\frac{y^2}{1/9}$. This implies $b^2 = 1/9$, so $b = \sqrt{1/9} = 1/3$.

Substituting these values, we obtain the standard form of an ellipse: $\frac{(x - 2)^2}{1^2} + \frac{y^2}{\left(\frac{1}{3}\right)^2} = 1$ From this standard form, we can identify the key properties of the ellipse:

• The center of the ellipse is $(h, k) = (2, 0)$.
• The semi-axis length along the $x$-direction is $a = 1$.
• The semi-axis length along the $y$-direction is $b = 1/3$.

4. Determine the Range of x and y For any real values $x, y$ that satisfy this ellipse equation, the terms on the left side must be non-negative. Since their sum is 1, each term must individually be less than or equal to 1.

• Range for x: The term involving $x$ is $\frac{(x - 2)^2}{1^2}$. For this to be part of an ellipse equation equal to 1, it must satisfy: $\frac{(x - 2)^2}{1^2} \le 1$ $(x - 2)^2 \le 1$ Taking the square root of both sides, we must remember the property $\sqrt{u^2} = |u|$: $|x - 2| \le 1$ This absolute value inequality means that $x-2$ must lie between $-1$ and $1$, inclusive: $-1 \le x - 2 \le 1$ To isolate $x$, we add 2 to all parts of the inequality: $-1 + 2 \le x \le 1 + 2$ $1 \le x \le 3$ Therefore, $x \in [1, 3]$.

• Range for y: Similarly, the term involving $y$ is $\frac{y^2}{\left(\frac{1}{3}\right)^2}$. This term must also be less than or equal to 1: $\frac{y^2}{\left(\frac{1}{3}\right)^2} \le 1$ Multiply both sides by $\left(\frac{1}{3}\right)^2$: $y^2 \le \left(\frac{1}{3}\right)^2$ Taking the square root of both sides: $|y| \le \frac{1}{3}$ This inequality means that $y$ must lie between $-1/3$ and $1/3$, inclusive: $-\frac{1}{3} \le y \le \frac{1}{3}$ Therefore, $y \in \left[ -\frac{1}{3}, \frac{1}{3} \right]$.

5. Compare with Options Our derived ranges are:

• $x \in [1, 3]$
• $y \in \left[ -\frac{1}{3}, \frac{1}{3} \right]$

Let's check the given options: (A) $\left[ - \frac{1}{3}, \frac{1}{3} \right]$ and $\left[ - \frac{1}{3}, \frac{1}{3} \right]$ (B) $\left[ - \frac{1}{3}, \frac{1}{3} \right]$ and $[1, 3]$ (C) $[1, 3]$ and $[1, 3]$ (D) $[1, 3]$ and $\left[ - \frac{1}{3}, \frac{1}{3} \right]$

Our calculated ranges for $x$ and $y$ perfectly match Option (D).

---

Important Tips and Common Mistakes:

1. Completing the Square with a Coefficient: If you have an expression like $Ax^2 + Bx$, first factor out $A$: $A(x^2 + (B/A)x)$. Then complete the square inside the parenthesis: $A \left( (x + \frac{B}{2A})^2 - \left(\frac{B}{2A}\right)^2 \right)$. Remember to distribute $A$ back or adjust the constant term carefully. In this problem, the coefficient of $x^2$ was 1, simplifying this step.
2. Correctly Identifying Denominators for Ellipse Standard Form: When you have an equation like $C(y-k)^2 = D$, to get it into the standard form $\frac{(y-k)^2}{b^2} = 1$, you must divide by $D$: $\frac{C(y-k)^2}{D} = 1 \implies \frac{(y-k)^2}{D/C} = 1$. So, $b^2 = D/C$. A common error is to directly take $b^2 = C$ or $b^2 = D$. In our problem, $9y^2 = 1$, so $y^2 = 1/9$, which correctly gives $b^2 = 1/9$ and $b = 1/3$.
3. Deriving Ranges from Absolute Value: Always remember that $\sqrt{u^2} = |u|$. For an inequality like $|X-h| \le a$, it translates to $-a \le X-h \le a$. This is a crucial step to correctly determine the bounds of the interval. Don't forget the negative bound!
4. Checking for Real Solutions: Before determining ranges, ensure that the equation results in a valid conic section. If, for instance, the right-hand side of the standard form were negative, or if it were zero (representing a single point or a line), the interpretation of ranges would be different or no real solution would exist. Here, the right-hand side is 1, confirming an ellipse.

---

Key Takeaway: Problems involving general quadratic equations in two variables, especially when asking for the range of these variables, are often solved by converting the equation into the standard form of a conic section (ellipse, parabola, hyperbola, or degenerate cases). The technique of completing the square is the indispensable algebraic tool for this transformation. Once in standard form, the geometric properties (like center and semi-axes for an ellipse) directly provide the bounds for the variables. Always perform algebraic manipulations carefully and understand the interpretation of the standard forms.