The locus of the foot of perpendicular drawn from the centre of the ellipse `x^2+""3y^2=""6` on any tangent to it is (1) `(x^2-y^2)^2=""6x^2+""2y^2` (2) `(x^2-y^2)^2=""6x^2-2y^2` (3) `(x^2+y^2)^2=""6x^2+""2y^2` (4) `(x^2+y^2)^2=""6x^2-2y^2`

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 \(x^2 + 3y^2 = 6\) to any tangent to the ellipse, we will follow these steps:

### Step 1: Rewrite the equation of the ellipse
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 is in the standard form of an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a^2 = 6\) and \(b^2 = 2\).

### Step 2: Identify the center of the ellipse
The center of the ellipse is at the origin \((0, 0)\).

### Step 3: General equation of the tangent to the ellipse
The general equation of the tangent to the ellipse at point \((x_1, y_1)\) on the ellipse is given by:
\[ \frac{xx_1}{6} + \frac{yy_1}{2} = 1 \]
where \((x_1, y_1)\) satisfies the ellipse equation.

### Step 4: Find the foot of the perpendicular from the center to the tangent
The foot of the perpendicular from the center \((0, 0)\) to the tangent line can be found using the formula for the foot of the perpendicular from a point to a line. The line can be rewritten as:
\[ \frac{xx_1}{6} + \frac{yy_1}{2} - 1 = 0 \]
Using the formula for the foot of the perpendicular from point \((x_0, y_0)\) to the line \(Ax + By + C = 0\):
\[ \left( \frac{B(Bx_0 - Ay_0) - AC}{A^2 + B^2}, \frac{A(Ay_0 - Bx_0) - BC}{A^2 + B^2} \right) \]
Substituting \(A = \frac{x_1}{6}\), \(B = \frac{y_1}{2}\), and \(C = -1\) with \((x_0, y_0) = (0, 0)\), we find the coordinates of the foot of the perpendicular.

### Step 5: Substitute and simplify
After substituting and simplifying, we will derive a relationship between \(x\) and \(y\) that represents the locus of the foot of the perpendicular.

### Step 6: Final equation
After performing the calculations, we arrive at the equation:
\[ (x^2 + y^2)^2 = 6x^2 + 2y^2 \]
This matches option (3).

### Conclusion
The locus of the foot of the perpendicular drawn from the center of the ellipse on any tangent to it is given by:
\[ (x^2 + y^2)^2 = 6x^2 + 2y^2 \]