Normal at `(2, 2)` to curve `x^2 + 2xy – 3y^2 = 0` is L. Then perpendicular distance from origin to line L is (A )4 `sqrt(2) ` (B) 2 (C) 2`sqrt(2) ` (D) 4

Options

A. `4sqrt(2)`
B. `2`
C. `2sqrt(2)`
D. 4

Explanation

To solve the problem, we need to find the equation of the normal to the curve \(x^2 + 2xy - 3y^2 = 0\) at the point \((2, 2)\) and then calculate the perpendicular distance from the origin to this normal line.

### Step 1: Differentiate the curve to find the slope of the tangent
We start with the implicit differentiation of the curve \(x^2 + 2xy - 3y^2 = 0\).

Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(2xy) - \frac{d}{dx}(3y^2) = 0 \]
This gives:
\[ 2x + 2\left(y + x\frac{dy}{dx}\right) - 6y\frac{dy}{dx} = 0 \]
Rearranging this, we have:
\[ 2x + 2y + 2x\frac{dy}{dx} - 6y\frac{dy}{dx} = 0 \]
\[ 2x + 2y = (6y - 2x)\frac{dy}{dx} \]
Thus, we can express \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{2x + 2y}{6y - 2x} \]

### Step 2: Calculate the slope of the tangent at the point (2, 2)
Substituting \(x = 2\) and \(y = 2\):
\[ \frac{dy}{dx} = \frac{2(2) + 2(2)}{6(2) - 2(2)} = \frac{4 + 4}{12 - 4} = \frac{8}{8} = 1 \]
The slope of the tangent at the point \((2, 2)\) is \(1\).

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{1} = -1 \]

### Step 4: 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 \(m = -1\):
\[ y - 2 = -1(x - 2) \]
This simplifies to:
\[ y - 2 = -x + 2 \implies y = -x + 4 \]
Rearranging gives:
\[ x + y - 4 = 0 \]

### Step 5: Calculate the perpendicular distance from the origin to the normal line
The formula for the perpendicular distance \(d\) from a point \((\alpha, \beta)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|A\alpha + B\beta + C|}{\sqrt{A^2 + B^2}} \]
For our line \(x + y - 4 = 0\), we have \(A = 1\), \(B = 1\), and \(C = -4\). The origin is \((0, 0)\):
\[ d = \frac{|1(0) + 1(0) - 4|}{\sqrt{1^2 + 1^2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

### Final Answer
The perpendicular distance from the origin to the line \(L\) is \(2\sqrt{2}\).