The locus of the foot of prependicular drawn from the center of the ellipse `x^(2)+3y^(2)=6` on any tangent to it is

Options

A. `(x^(2)+y^(2))^(2)=6x^(2)+2y^(2)`
B. `(x^(2)+y^(2))^(2)=6x^(2)-2y^(2)`
C. `(x^(2)-y^(2))^(2)=6x^(2)+2y^(2)`
D. `(x^(2)-y^(2))^(2)=6x^(2)-2y^(2)`

Answer

To find the locus of the foot of the perpendicular drawn from the center of the ellipse \( x^2 + 3y^2 = 6 \) on any tangent to it, we can follow these steps:

### Step 1: Write the equation of the ellipse in standard form
The given equation of the ellipse is:

\[ x^2 + 3y^2 = 6 \]

Dividing the entire equation by 6, we get:

\[ \frac{x^2}{6} + \frac{y^2}{2} = 1 \]

This shows that \( a^2 = 6 \) and \( b^2 = 2 \).

### Step 2: Write the equation of the tangent to the ellipse
The equation of the tangent to the ellipse at point \( (x_1, y_1) \) is given by:

\[ \frac{xx_1}{6} + \frac{yy_1}{2} = 1 \]

### Step 3: Find the slope of the tangent
The slope of the tangent line can be expressed as:

\[ y = mx + \sqrt{a^2m^2 + b^2} \]

Substituting \( a^2 = 6 \) and \( b^2 = 2 \):

\[ y = mx + \sqrt{6m^2 + 2} \]

### Step 4: Find the equation of the perpendicular from the center
The center of the ellipse is at the origin (0, 0). The equation of the line perpendicular to the tangent at the origin can be expressed as:

\[ y - 0 = -\frac{1}{m}(x - 0) \quad \Rightarrow \quad y = -\frac{1}{m}x \]

### Step 5: Substitute to find the foot of the perpendicular
To find the foot of the perpendicular, we need to solve the equations of the tangent and the perpendicular simultaneously. Substituting \( y = -\frac{1}{m}x \) into the tangent equation:

\[ -\frac{1}{m}x = mx + \sqrt{6m^2 + 2} \]

### Step 6: Solve for \( x \)
Rearranging gives:

\[ -\frac{1}{m}x - mx = \sqrt{6m^2 + 2} \]

Factoring out \( x \):

\[ x\left(-\frac{1}{m} - m\right) = \sqrt{6m^2 + 2} \]

Thus,

\[ x = \frac{\sqrt{6m^2 + 2}}{-\frac{1}{m} - m} \]

### Step 7: Find \( y \)
Substituting \( x \) back into \( y = -\frac{1}{m}x \):

\[ y = -\frac{1}{m}\left(\frac{\sqrt{6m^2 + 2}}{-\frac{1}{m} - m}\right) \]

### Step 8: Find the locus
To find the locus, we eliminate \( m \) from the equations of \( x \) and \( y \). This involves some algebraic manipulation to express \( x \) and \( y \) in terms of each other.

After simplification, we arrive at the equation:

\[ x^2 + y^2 = \sqrt{6x^2 + 2y^2} \]

Squaring both sides gives:

\[ (x^2 + y^2)^2 = 6x^2 + 2y^2 \]

This is the required locus.