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**Hint:**Here, we will be using the concept of finding the slope of the normal to any curve and the formula for equation of any line which slope m and passing through a point.

Given equation of the curve is ${x^2} + 2xy - 3{y^2} = 0{\text{ }} \to {\text{(1)}}$

**Complete step-by-step answer:**

Here, we need to first of all find the equation of the normal to the given curve. As, we know that the slope of normal to any curve is $ - \dfrac{{dx}}{{dy}}$ .

Let us differentiate the given equation of curve both sides with respect to x, we get

\[

\dfrac{{d\left( {{x^2} + 2xy - 3{y^2}} \right)}}{{dx}} = 0 \Rightarrow \dfrac{{d\left( {{x^2}} \right)}}{{dx}} + 2\left[ {\dfrac{{d\left( {xy} \right)}}{{dx}}} \right] - 3\left[ {\dfrac{{d\left( {{y^2}} \right)}}{{dx}}} \right] = 0 \Rightarrow 2x + 2\left[ {y\left( {\dfrac{{dx}}{{dx}}} \right) + x\left( {\dfrac{{dy}}{{dx}}} \right)} \right] - 3 \times 2y\left[ {\dfrac{{dy}}{{dx}}} \right] = 0 \\

\Rightarrow 2x + 2y + 2x\left( {\dfrac{{dy}}{{dx}}} \right) - 6y\left( {\dfrac{{dy}}{{dx}}} \right) = 0 \Rightarrow 2\left( {x + y} \right) + 2\left( {\dfrac{{dy}}{{dx}}} \right)\left( {x - 3y} \right) = 0 \Rightarrow 2\left( {\dfrac{{dy}}{{dx}}} \right)\left( {x - 3y} \right) = - 2\left( {x + y} \right) \\

\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = - \dfrac{{2\left( {x + y} \right)}}{{2\left( {x - 3y} \right)}} \Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) = - \dfrac{{\left( {x + y} \right)}}{{\left( {x - 3y} \right)}} \\

\]

Slope of normal to the given curve$ = - \dfrac{{dx}}{{dy}} = \dfrac{{ - 1}}{{\dfrac{{dy}}{{dx}}}} = \dfrac{{ - 1}}{{\left[ { - \dfrac{{\left( {x + y} \right)}}{{\left( {x - 3y} \right)}}} \right]}} = \dfrac{{\left( {x - 3y} \right)}}{{\left( {x + y} \right)}}$

In order to find the slope of the normal to the given curve at a point $\left[ {1,1} \right]$, let us substitute x=1 and y=1 in the above expression of slope.

Slope of normal to the given curve at $\left[ {1,1} \right]$$ = - \dfrac{{dx}}{{dy}} = \dfrac{{\left( {1 - 3 \times 1} \right)}}{{\left( {1 + 1} \right)}} = \dfrac{{1 - 3}}{2} = - 1$

Also, we know that equation of the line with slope m and passing through any point (${x_1},{y_1}$) is given by $y - {y_1} = m\left( {x - {x_1}} \right)$

Now, equation of the normal with slope of -1 and passing through point $\left[ {1,1} \right]$ is given by

$y - 1 = - 1\left( {x - 1} \right) \Rightarrow y - 1 = - x + 1 \Rightarrow x + y - 1 - 1 = 0 \Rightarrow x + y - 2 = 0{\text{ }} \to {\text{(2)}}$

Now, in order to find out whether this normal meets the given curve at any point other than the given point $\left[ {1,1} \right]$. Let us solve for the common solutions between the curve and the normal.

Equation (2) can be written as $ \Rightarrow x + y - 2 = 0 \Rightarrow x = 2 - y{\text{ }} \to {\text{(3)}}$

Substitute equation (3) in equation (1), we get

${\left( {2 - y} \right)^2} + 2y\left( {2 - y} \right) - 3{y^2} = 0 \Rightarrow 4 + {y^2} - 4y + 4y - 2{y^2} - 3{y^2} = 0 \Rightarrow 4 - 4{y^2} = 0 \Rightarrow {y^2} = 1 \Rightarrow y = \pm 1$

When y=1, equation (3) reduces to $ \Rightarrow x = 2 - y = 2 - 1 \Rightarrow x = 1$

When y=-1, equation (3) reduces to $ \Rightarrow x = 2 - y = 2 - \left( { - 1} \right) \Rightarrow x = 3$

So, the points where the given curve meets with the normal to this curve are (1,1) and (3,-1).

Clearly, apart from point (1,1) the given curve meets the normal to this curve at (3,-1) which lies in the fourth quadrant because in the fourth quadrant x-coordinate is positive and y-coordinate is negative.

**Hence, option D is correct.**

**Note:**In these types of problems, we will simply find the equation for the normal to the given curve at the given point by finding the slope of the normal. Then, we will find the points of intersection of the given curve and normal to the given curve and if any other point appears except the given point that means that the normal meets the curve again else it does not.

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