Answer

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**Hint:**

Here, we have to find the point where the normal to the curve cuts the curve again. First we will find the slope of the tangent, and then we will find the slope of the normal. We will use the equation of line with one point and slope to find the equation of the normal, then we will find the point which cuts the normal again by substituting these points in the equation of the normal.

**Formula Used:**

We will use the following formula:

1) Slope of two perpendicular lines is given by \[{m_1}{m_2} = - 1\]

2) Equation of line with one point and slope formula is given by \[y - {y_1} = m\left( {x - {x_1}} \right)\]

**Complete step by step solution:**

Let \[\left( {x,y} \right)\] be the point where the normal meets the curve again.

We are given the equation of a curve \[2{x^2} + {y^2} = 12\].

Now, we will find the slope of the curve by differentiating the given equation

Differentiating \[2{x^2} + {y^2} = 12\] with respect to \[x\], we get

\[ \Rightarrow 4x + 2y\dfrac{{dy}}{{dx}} = 0\]

Rewriting the equation, we get

\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{4x}}{{2y}}\]

\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{2x}}{y}\]

Now, we will find the slope at the point \[\left( {2,2} \right)\] by substituting these values, we get

\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {2,2} \right)}} = - \dfrac{{2\left( 2 \right)}}{2}\]

\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{\left( {2,2} \right)}} = - 2\]

Since the tangent line and normal line are perpendicular lines, we use the condition to find the slope of the normal.

Slope of two perpendicular lines is given by \[{m_1}{m_2} = - 1\]

\[ \Rightarrow - 2 \cdot {m_2} = - 1\]

\[ \Rightarrow {m_2} = \dfrac{1}{2}\]

Thus the slope of the normal is \[\dfrac{1}{2}\].

Now, we will find the equation of the normal by using the equation of a line with one point and slope formula.

We are given the point \[\left( {2,2} \right)\]and slope \[\dfrac{1}{2}\].

Equation of line with one point and slope formula is given by \[y - {y_1} = m\left( {x - {x_1}} \right)\].

Substituting \[{y_1} = 2\], \[{x_1} = 2\] and \[m = \dfrac{1}{2}\] in the slope formula, we get

\[y - 2 = \dfrac{1}{2}\left( {x - 2} \right)\]

Multiplying and rewriting the equation,

\[ \Rightarrow 2\left( {y - 2} \right) = \left( {x - 2} \right)\]

\[ \Rightarrow 2y - 4 = x - 2\]

Subtracting \[x\] from both sides, we get

\[ \Rightarrow 2y - 4 - x = - 2\]

Adding 4 on both sides, we get

\[ \Rightarrow 2y - x = 2\]

Thus, the equation of the normal is \[2y - x = 2\].

Now, we will substitute the points given in the option in the equation of normal.

Now, Substituting \[\left( { - \dfrac{{22}}{9}, - \dfrac{2}{9}} \right)\] in \[2y - x = 2\], we get

\[\begin{array}{l} \Rightarrow 2y - x = 2\\ \Rightarrow 2\left( {\dfrac{{ - 2}}{9}} \right) - \left( {\dfrac{{ - 22}}{9}} \right) = 2\end{array}\]

Taking LCM, we get

\[ \Rightarrow \dfrac{{ - 4 + 22}}{9} = 2\]

On cross multiplication, we get

\[ \Rightarrow - 4 + 22 = 2 \cdot 9\]

\[ \Rightarrow 18 = 18\]

Thus, the point \[\left( { - \dfrac{{22}}{9}, - \dfrac{2}{9}} \right)\] satisfies the condition.

Now, Substituting \[\left( {\dfrac{{22}}{9},\dfrac{2}{9}} \right)\] in \[2y - x = 2\], we get

\[ \Rightarrow 2\left( {\dfrac{2}{9}} \right) - \left( {\dfrac{{22}}{9}} \right) = 2\]

Taking LCM, we get

\[ \Rightarrow \dfrac{{4 - 22}}{9} = 2\]

On cross multiplication, we get

\[ \Rightarrow 4 - 22 = 2 \cdot 9\]

\[ \Rightarrow - 18 \ne 18\]

Thus, the point \[\left( {\dfrac{{22}}{9},\dfrac{2}{9}} \right)\] does not satisfy the condition.

Now, Substituting \[\left( { - 2, - 2} \right)\] in \[2y - x = 2\], we get

\[ \Rightarrow 2\left( { - 2} \right) - \left( { - 2} \right) = 2\]

Multiplying the terms, we get

\[\begin{array}{l} \Rightarrow - 4 + 4 = 2\\ \Rightarrow 0 = 2\end{array}\]

\[ \Rightarrow 0 \ne 2\]

Thus, the point \[\left( { - 2, - 2} \right)\] does not satisfy the condition.

Therefore, the point where the normal to the curve cuts the curve again is \[\left( { - \dfrac{{22}}{9}, - \dfrac{2}{9}} \right)\].

**Hence, Option (A) is correct.**

**Note:**

We should know that the normal is a line that is perpendicular to any object. Here, we have the normal as the line perpendicular to the tangent of the curve. We should also know that the slope of the line can be found by differentiating the equation of the line. Since the normal and the tangent line are perpendicular, so the slope of the normal is the negative reciprocal of the slope of the tangent.

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