Question

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

• A. $[1, 3]$ and $[1, 3]$
• B. $[1, 3]$ and $\left[-\frac{1}{3}, \frac{1}{3}\right]$
• C. $\left[-\frac{1}{3}, \frac{1}{3}\right]$ and $[1, 3]$
• D. $\left[-\frac{1}{3}, \frac{1}{3}\right]$ and $\left[-\frac{1}{3}, \frac{1}{3}\right]$

Hint:

For any equation of the form $(x-h)^2 + k(y-j)^2 = C$, the maximum and minimum values are found by setting one square term to zero and solving for the other.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given equation represents an ellipse. Completing the square will help identify the range of values for $x$ and $y$.
Step 2: Detailed Explanation:
Given: $x^2 - 4x + 9y^2 + 3 = 0$.
Complete the square for $x$:
\[ (x^2 - 4x + 4) + 9y^2 + 3 - 4 = 0 \]
\[ (x-2)^2 + 9y^2 = 1 \]
This is a standard ellipse: $\frac{(x-2)^2}{1} + \frac{y^2}{(1/3)^2} = 1$.
For real $x, y$:
1. $(x-2)^2 \leq 1 \implies -1 \leq x-2 \leq 1 \implies 1 \leq x \leq 3$.
2. $9y^2 \leq 1 \implies y^2 \leq \frac{1}{9} \implies -\frac{1}{3} \leq y \leq \frac{1}{3}$.
Step 3: Final Answer:
$x \in [1, 3]$ and $y \in \left[-\frac{1}{3}, \frac{1}{3}\right]$.