The normal to the curve `2x^2+y^2=12` at the point `(2,2)` cuts the curve again at (a) `(-(22)/9,-2/9)` (b) `((22)/9,2/9)` `(-2,-2)` (d) none of these

Options

A. `(-(22)/(9),-(2)/(9))`
B. `((22)/(9),(2)/(9))`
C. `(-2,-2)`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Differentiate the given equation
We start with the equation of the curve:
\[ 2x^2 + y^2 = 12 \]

We differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(2x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(12) \]
This gives us:
\[ 4x + 2y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ 2y \frac{dy}{dx} = -4x \]
\[ \frac{dy}{dx} = -\frac{2x}{y} \]

### Step 3: Find the slope of the tangent at the point (2, 2)
Substituting the point \( (2, 2) \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{(2, 2)} = -\frac{2(2)}{2} = -2 \]
Thus, the slope of the tangent at the point \( (2, 2) \) is \(-2\).

### Step 4: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{-2} = \frac{1}{2} \]

### Step 5: Write the equation of the normal line
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (2, 2) \) and \( m = \frac{1}{2} \):
\[ y - 2 = \frac{1}{2}(x - 2) \]
Simplifying this, we get:
\[ y - 2 = \frac{1}{2}x - 1 \]
\[ y = \frac{1}{2}x + 1 \]

### Step 6: Substitute the normal line equation into the curve equation
We substitute \(y = \frac{1}{2}x + 1\) into the curve equation \(2x^2 + y^2 = 12\):
\[ 2x^2 + \left(\frac{1}{2}x + 1\right)^2 = 12 \]
Expanding the equation:
\[ 2x^2 + \left(\frac{1}{4}x^2 + x + 1\right) = 12 \]
Combining like terms:
\[ 2x^2 + \frac{1}{4}x^2 + x + 1 = 12 \]
\[ \frac{8}{4}x^2 + \frac{1}{4}x^2 + x + 1 = 12 \]
\[ \frac{9}{4}x^2 + x + 1 - 12 = 0 \]
\[ \frac{9}{4}x^2 + x - 11 = 0 \]

### Step 7: Multiply through by 4 to eliminate the fraction
\[ 9x^2 + 4x - 44 = 0 \]

### Step 8: Solve the quadratic equation using the quadratic formula
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
Here, \(a = 9\), \(b = 4\), and \(c = -44\):
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 9 \cdot (-44)}}{2 \cdot 9} \]
\[ x = \frac{-4 \pm \sqrt{16 + 1584}}{18} \]
\[ x = \frac{-4 \pm \sqrt{1600}}{18} \]
\[ x = \frac{-4 \pm 40}{18} \]
Calculating the two possible values for \(x\):
1. \(x = \frac{36}{18} = 2\) (the original point)
2. \(x = \frac{-44}{18} = -\frac{22}{9}\)

### Step 9: Find the corresponding \(y\) value
Substituting \(x = -\frac{22}{9}\) back into the normal line equation:
\[ y = \frac{1}{2}\left(-\frac{22}{9}\right) + 1 = -\frac{11}{9} + 1 = -\frac{11}{9} + \frac{9}{9} = -\frac{2}{9} \]

### Final Result
Thus, the normal to the curve at the point \( (2, 2) \) cuts the curve again at:
\[ \left(-\frac{22}{9}, -\frac{2}{9}\right) \]

### Answer
The correct option is (a) \(\left(-\frac{22}{9}, -\frac{2}{9}\right)\).