This problem asks us to find the locus of a specific point: the foot of the perpendicular drawn from the center of a given ellipse to any of its tangent lines. This particular locus is a well-known curve in geometry called the pedal curve of the ellipse with respect to its center.

1. Key Concept: The Pedal Curve of an Ellipse

For an ellipse centered at the origin with the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the locus of the foot of the perpendicular drawn from its center $(0,0)$ to any tangent is given by the equation: $\left( {{x^2} + {y^2}} \right) ^2 = a^2{x^2} + b^2{y^2}$ This equation is a powerful shortcut if remembered. However, for a complete understanding and for cases where such a formula is not readily available, we will derive it step-by-step using the general strategy for locus problems.

General Strategy for Locus Problems:

1. Assume the coordinates of the moving point (whose locus we need to find) are $(x, y)$ (or $(h, k)$).
2. Formulate equations based on the given geometric conditions. These equations will typically involve the coordinates $(x, y)$ and one or more parameters.
3. Eliminate these parameters to obtain a single equation relating $x$ and $y$. This final equation represents the locus.

Let's apply this strategy to our problem.

2. Step 1: Standardizing the Ellipse Equation

The given equation of the ellipse is: $x^2 + 3y^2 = 6$ To use standard formulas for tangents and to identify $a^2$ and $b^2$, we must convert this into the standard form of an ellipse, which is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Why this step is crucial: The standard form clearly reveals the values of $a^2$ and $b^2$, which are fundamental parameters for the ellipse and are used in tangent equations and the general pedal curve formula.

To achieve the standard form, divide the entire equation by 6: $\frac{x^2}{6} + \frac{3y^2}{6} = \frac{6}{6}$ $\frac{x^2}{6} + \frac{y^2}{2} = 1$

By comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we can identify the values of $a^2$ and $b^2$: $a^2 = 6 \quad \text{and} \quad b^2 = 2$ The center of this ellipse is at the origin $(0,0)$, which is exactly the point from which the perpendicular is drawn.

3. Step 2: Formulating the Equation of Any Tangent to the Ellipse

Next, we need a general representation for any tangent to the ellipse. The most convenient form for this problem, as it involves the slope and the center, is the slope-intercept form of the tangent.

Why this step is crucial: The foot of the perpendicular, whose locus we seek, lies on the tangent line. Therefore, we need an equation that describes all possible tangent lines.

The equation of any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is given by the formula: $y = mx \pm \sqrt{a^2m^2 + b^2} \quad \ldots(1)$ Substitute the values $a^2=6$ and $b^2=2$ into this equation: $y = mx \pm \sqrt{6m^2 + 2}$ Here, $m$ is a parameter representing the slope of the tangent. We will need to eliminate this parameter later to find the locus. The $\pm$ sign indicates that for a given slope $m$, there are two parallel tangents to the ellipse.

4. Step 3: Formulating the Equation of the Perpendicular from the Center

We are looking for the locus of the foot of the perpendicular drawn from the center of the ellipse, which is $(0,0)$, to any tangent. Let $(x, y)$ be this foot of the perpendicular.

Why this step is crucial: The point $(x,y)$ also lies on the line segment connecting the center to the foot of the perpendicular. This line is perpendicular to the tangent.

• The slope of the tangent line is $m$.
• A line perpendicular to the tangent will have a slope of $-\frac{1}{m}$ (provided $m \ne 0$). This is a fundamental property of perpendicular lines: the product of their slopes is $-1$.
• This perpendicular line passes through the center $(0,0)$.

Using the point-slope form of a line, $Y - Y_1 = \text{slope}(X - X_1)$, the equation of this perpendicular line is: $y - 0 = \left(-\frac{1}{m}\right)(x - 0)$ $y = -\frac{x}{m} \quad \ldots(2)$ This equation defines the line segment from the origin $(0,0)$ to the foot of the perpendicular $(x,y)$.

5. Step 4: Eliminating the Parameter 'm' to Find the Locus

The point $(x, y)$ whose locus we seek is the foot of the perpendicular. This point lies both on the tangent line (Equation 1) and on the perpendicular line from the center (Equation 2). Therefore, $(x, y)$ must satisfy both equations simultaneously. Our goal is to find a relationship between $x$ and $y$ that does not involve the parameter $m$.

Why this step is crucial: This is the core of any locus problem. By satisfying both conditions, the point $(x,y)$ is uniquely defined for a given parameter $m$. Eliminating $m$ gives us the general path traced by $(x,y)$.

From Equation (2), we can express $m$ in terms of $x$ and $y$: $my = -x$ $m = -\frac{x}{y} \quad \text{(assuming } y \ne 0 \text{)}$ Why this manipulation: We need to substitute $m$ into Equation (1). Expressing $m$ in terms of $x$ and $y$ allows us to eliminate the parameter.

Now, substitute this expression for $m$ into Equation (1), the tangent equation: $y = \left(-\frac{x}{y}\right)x \pm \sqrt{6\left(-\frac{x}{y}\right)^2 + 2}$ $y = -\frac{x^2}{y} \pm \sqrt{6\frac{x^2}{y^2} + 2}$

To simplify, we want to isolate the square root term before squaring. This is a common algebraic technique to handle square roots and the $\pm$ sign efficiently. Move the term $-\frac{x^2}{y}$ to the left side: $y + \frac{x^2}{y} = \pm \sqrt{\frac{6x^2}{y^2} + \frac{2y^2}{y^2}}$ Combine the terms on the left side into a single fraction and simplify the expression under the radical on the right side: $\frac{y^2 + x^2}{y} = \pm \sqrt{\frac{6x^2 + 2y^2}{y^2}}$ $\frac{x^2 + y^2}{y} = \pm \frac{\sqrt{6x^2 + 2y^2}}{|y|}$ Note that $\sqrt{y^2} = |y|$.

To eliminate the $\pm$ sign and the square root, we square both sides of the equation. $\left(\frac{x^2 + y^2}{y}\right)^2 = \left(\pm \frac{\sqrt{6x^2 + 2y^2}}{|y|}\right)^2$ $\frac{(x^2 + y^2)^2}{y^2} = \frac{6x^2 + 2y^2}{y^2}$

Since the foot of the perpendicular $(x,y)$ cannot be the origin $(0,0)$ if $y=0$ (as the tangent cannot pass through the origin in general, unless it's a degenerate case which is not relevant here), we can safely multiply both sides by $y^2$: $(x^2 + y^2)^2 = 6x^2 + 2y^2$ This is the required locus of the foot of the perpendicular. This derived equation perfectly matches the general formula $(x^2 + y^2)^2 = a^2x^2 + b^2y^2$ with $a^2=6$ and $b^2=2$.

Consideration for $y=0$ (Edge Case): If $y=0$, then from Equation (2), $y = -x/m$ implies $x=0$ (assuming $m$ is finite). So, the only point on the x-axis that can be the foot of the perpendicular is the origin $(0,0)$. Let's check if $(0,0)$ satisfies the derived locus equation: $(0^2+0^2)^2 = 6(0)^2+2(0)^2 \Rightarrow 0=0$. Thus, the origin is indeed included in the locus, and our derivation (assuming $y \ne 0$ for division) holds for all points on the locus. Similarly, if $m$ were undefined (vertical tangent $x=\pm a$), then $y = -x/m$ would imply $x=0$, leading to $(0,0)$. If $m=0$ (horizontal tangent $y=\pm b$), then $x=0$, again leading to $(0,0)$. The derivation using $m$ handles all cases implicitly.

6. Step 5: Comparing with Options and Final Answer

The derived locus equation is: $(x^2 + y^2)^2 = 6x^2 + 2y^2$

Comparing this with the given options: (A) $\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} + 2{y^2}$ (B) $$\