The length of the perpendicular from the origin, on the normal to the curve, `x^2 + 3xy - 10 y^2 = 0` at the point (2,1) is :

Options

A. `3 sqrt5`
B. `3`
C. `sqrt5`
D. `2/sqrt5`

Explanation

To solve the problem, we need to find the length of the perpendicular from the origin to the normal of the curve \(x^2 + 3xy - 10y^2 = 0\) at the point (2, 1). Here are the steps to arrive at the solution:

### Step 1: Differentiate the curve implicitly
We start with the equation of the curve:
\[ x^2 + 3xy - 10y^2 = 0 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(3xy) - \frac{d}{dx}(10y^2) = 0 \]
This gives:
\[ 2x + 3\left(y + x\frac{dy}{dx}\right) - 20y\frac{dy}{dx} = 0 \]

### Step 2: Rearranging the equation
Rearranging the equation:
\[ 2x + 3y + 3x\frac{dy}{dx} - 20y\frac{dy}{dx} = 0 \]
Combining the terms with \(\frac{dy}{dx}\):
\[ 3x\frac{dy}{dx} - 20y\frac{dy}{dx} = - (2x + 3y) \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(3x - 20y) = - (2x + 3y) \]
Thus,
\[ \frac{dy}{dx} = \frac{-(2x + 3y)}{3x - 20y} \]

### Step 3: Calculate the slope at the point (2, 1)
Substituting \(x = 2\) and \(y = 1\):
\[ \frac{dy}{dx} = \frac{-(2(2) + 3(1))}{3(2) - 20(1)} = \frac{-(4 + 3)}{6 - 20} = \frac{-7}{-14} = \frac{1}{2} \]

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

### Step 5: Write the equation of the normal
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = -2\), \(x_1 = 2\), and \(y_1 = 1\):
\[ y - 1 = -2(x - 2) \]
This simplifies to:
\[ y - 1 = -2x + 4 \implies y = -2x + 5 \]

### Step 6: Convert to standard form
Rearranging gives:
\[ 2x + y - 5 = 0 \]
Thus, \(a = 2\), \(b = 1\), and \(c = -5\).

### Step 7: Find the length of the perpendicular from the origin
The formula for the distance \(d\) from a point \((x_0, y_0)\) to the line \(ax + by + c = 0\) is:
\[ d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \]
Substituting \(x_0 = 0\), \(y_0 = 0\):
\[ d = \frac{|2(0) + 1(0) - 5|}{\sqrt{2^2 + 1^2}} = \frac{|-5|}{\sqrt{4 + 1}} = \frac{5}{\sqrt{5}} = \sqrt{5} \]

### Final Answer
The length of the perpendicular from the origin to the normal at the point (2, 1) is:
\[ \sqrt{5} \]