1. Write the ellipse in standard form

Given ellipse: $x^2+3y^2=6$ Divide by $6$: $\frac{x^2}{6}+\frac{y^2}{2}=1$ So, $a^2=6,\qquad b^2=2$ Its centre is the origin $(0,0)$.

2. Equation of tangent to the ellipse

A tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at point $(x_1,y_1)$ on the ellipse is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$ Here, $\frac{xx_1}{6}+\frac{yy_1}{2}=1$ Multiply by $6$: $xx_1+3yy_1=6$ So the tangent is $x_1x+3y_1y-6=0$ where $(x_1,y_1)$ lies on the ellipse: $x_1^2+3y_1^2=6$

3. Foot of perpendicular from the centre to this tangent

For line $Ax+By+C=0$ the foot of perpendicular from origin $(0,0)$ is $\left(\frac{-AC}{A^2+B^2},\frac{-BC}{A^2+B^2}\right)$

Here, $A=x_1,\quad B=3y_1,\quad C=-6$ Hence foot $P(x,y)$ is $x=\frac{6x_1}{x_1^2+9y_1^2},\qquad y=\frac{18y_1}{x_1^2+9y_1^2}$

4. Relate $x,y$ with $x_1,y_1$

Let $D=x_1^2+9y_1^2$ Then $x=\frac{6x_1}{D},\qquad y=\frac{18y_1}{D}$ So, $x_1=\frac{xD}{6},\qquad y_1=\frac{yD}{18}$

Now use $D=x_1^2+9y_1^2$ Substitute: $D=\left(\frac{xD}{6}\right)^2+9\left(\frac{yD}{18}\right)^2$ $D=D^2\left(\frac{x^2}{36}+\frac{y^2}{36}\right)$ $D=D^2\cdot \frac{x^2+y^2}{36}$ Assuming $D\neq 0$, $1=D\cdot \frac{x^2+y^2}{36}$ $D=\frac{36}{x^2+y^2}$

5. Use the fact that $(x_1,y_1)$ lies on the ellipse

We know $x_1^2+3y_1^2=6$ Substitute $x_1=\frac{xD}{6},\ y_1=\frac{yD}{18}$: $\left(\frac{xD}{6}\right)^2+3\left(\frac{yD}{18}\right)^2=6$ $\frac{x^2D^2}{36}+\frac{3y^2D^2}{324}=6$ $\frac{x^2D^2}{36}+\frac{y^2D^2}{108}=6$ Multiply by $108$: $3x^2D^2+y^2D^2=648$ $D^2(3x^2+y^2)=648$ Now put $D=\frac{36}{x^2+y^2}$ so $D^2=\frac{1296}{(x^2+y^2)^2}$ Then $\frac{1296(3x^2+y^2)}{(x^2+y^2)^2}=648$ Divide by $648$: $\frac{2(3x^2+y^2)}{(x^2+y^2)^2}=1$ $2(3x^2+y^2)=(x^2+y^2)^2$ Therefore, $\boxed{(x^2+y^2)^2=6x^2+2y^2}$

6. Match with options

This is exactly Option A.

Verification with stored answer

Stored correct answer: A

My derived answer: A

So they agree.