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 :

Question

Accepted Answer

To determine the locus of the foot of the perpendicular drawn from the center of the ellipse $x^2 + 3y^2 = 6$ to any tangent to it, we need to follow a systematic approach.

### Explanation for Option 3:

1. **Equation of the Ellipse**: The given ellipse is $x^2 + 3y^2 = 6$. This can be rewritten in standard form as $\frac{x^2}{6} + \frac{y^2}{2} = 1$.

2. **Center of the Ellipse**: The center of this ellipse is at the origin $(0, 0)$.

3. **Equation of Tangent**: The equation of the tangent to the ellipse at any point $(x_1, y_1)$ on the ellipse is given by:
   $$
   \frac{xx_1}{6} + \frac{yy_1}{2} = 1
   $$

4. **Foot of the Perpendicular**: The foot of the perpendicular from the center $(0, 0)$ to this tangent line can be found using the formula for the perpendicular from a point to a line. The perpendicular distance from the origin to the tangent line is zero because the foot of the perpendicular lies on the line itself.

5. **Locus of the Foot of the Perpendicular**: To find the locus of the foot of the perpendicular, we consider the parametric coordinates of the point $(x_1, y_1)$ on the ellipse. Let $x_1 = \sqrt{6} \cos \theta$ and $y_1 = \sqrt{2} \sin \theta$.

6. **Substitute in Tangent Equation**: Substitute these into the tangent equation:
   $$
   \frac{x \cdot \sqrt{6} \cos \theta}{6} + \frac{y \cdot \sqrt{2} \sin \theta}{2} = 1
   $$
   Simplifying, we get:
   $$
   x \cos \theta + 3y \sin \theta = \sqrt{6}
   $$

7. **Foot of Perpendicular**: The foot of the perpendicular from the origin to this line is given by:
   $$
   \left( \frac{\sqrt{6} \cos \theta}{\cos^2 \theta + 9 \sin^2 \theta}, \frac{3\sqrt{6} \sin \theta}{\cos^2 \theta + 9 \sin^2 \theta} \right)
   $$

8. **Locus Equation**: Eliminating $\theta$ from these expressions, we derive the locus equation:
   $$
   (x^2 + y^2)^2 = 6x^2 + 2y^2
   $$

Thus, the correct locus of the foot of the perpendicular is given by Option 3: $(x^2 + y^2)^2 = 6x^2 + 2y^2$.

### Explanation for Incorrect Options:

- **Option 1**: $(x^2 - y^2)^2 = 6x^2 + 2y^2$ is incorrect because the derivation of the locus involves the sum of squares, not the difference.
  
- **Option 2**: $(x^2 - y^2)^2 = 6x^2 - 2y^2$ is incorrect for the same reason as Option 1, and additionally, the signs do not match the derived equation.

- **Option 4**: $(x^2 + y^2)^2 = 6x^2 - 2y^2$ is incorrect because the sign of the $y^2$ term does not match the derived equation.

In conclusion, the correct answer is Option 3, as it accurately represents the locus of the foot of the perpendicular from the center of the ellipse to any tangent line.