Let `P(x_0, y_0)` be the point on the hyperbola `3x^2 – 4y^2 = 36` which is nearest to line `3x +2y = 1`. Then `sqrt 2 (y_(0)-x_(0))` is equal to

Options

A. `-9`
B. `3`
C. `9`
D. `-3`

Answer

To solve the problem, we need to find the point \( P(x_0, y_0) \) on the hyperbola \( 3x^2 - 4y^2 = 36 \) that is nearest to the line \( 3x + 2y = 1 \). We will then compute \( \sqrt{2}(y_0 - x_0) \).

### Step 1: Rewrite the Hyperbola Equation
First, we rewrite the hyperbola equation in standard form:
\[ \frac{x^2}{12} - \frac{y^2}{9} = 1 \]
This shows that the hyperbola opens horizontally.

### Step 2: Find the Slope of the Line
Next, we find the slope of the line \( 3x + 2y = 1 \):
\[ 2y = -3x + 1 \implies y = -\frac{3}{2}x + \frac{1}{2} \]
Thus, the slope \( m \) of the line is \( -\frac{3}{2} \).

### Step 3: Use the Distance Formula
The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For our line \( 3x + 2y - 1 = 0 \), we have \( A = 3 \), \( B = 2 \), and \( C = -1 \). Thus, the distance becomes:
\[ d = \frac{|3x_0 + 2y_0 - 1|}{\sqrt{3^2 + 2^2}} = \frac{|3x_0 + 2y_0 - 1|}{\sqrt{13}} \]

### Step 4: Set Up the Lagrange Multiplier
To minimize the distance subject to the constraint of the hyperbola, we can use the method of Lagrange multipliers. We want to minimize:
\[ f(x_0, y_0) = |3x_0 + 2y_0 - 1| \]
subject to the constraint:
\[ g(x_0, y_0) = 3x_0^2 - 4y_0^2 - 36 = 0 \]

### Step 5: Solve Using Lagrange Multipliers
We set up the equations:
\[ \nabla f = \lambda \nabla g \]
Calculating the gradients:
\[ \nabla f = \left(3, 2\right), \quad \nabla g = \left(6x_0, -8y_0\right) \]
This gives us the system of equations:
1. \( 3 = \lambda (6x_0) \)
2. \( 2 = \lambda (-8y_0) \)

From the first equation, we can express \( \lambda \):
\[ \lambda = \frac{3}{6x_0} = \frac{1}{2x_0} \]
From the second equation:
\[ \lambda = \frac{2}{-8y_0} = -\frac{1}{4y_0} \]
Setting these equal gives:
\[ \frac{1}{2x_0} = -\frac{1}{4y_0} \implies 4y_0 = -2x_0 \implies 2y_0 = -x_0 \implies y_0 = -\frac{x_0}{2} \]

### Step 6: Substitute Back into the Hyperbola Equation
Substituting \( y_0 = -\frac{x_0}{2} \) into the hyperbola equation:
\[ 3x_0^2 - 4\left(-\frac{x_0}{2}\right)^2 = 36 \]
This simplifies to:
\[ 3x_0^2 - 4\left(\frac{x_0^2}{4}\right) = 36 \implies 3x_0^2 - x_0^2 = 36 \implies 2x_0^2 = 36 \implies x_0^2 = 18 \implies x_0 = 3\sqrt{2} \]
Then substituting back to find \( y_0 \):
\[ y_0 = -\frac{3\sqrt{2}}{2} \]

### Step 7: Calculate \( \sqrt{2}(y_0 - x_0) \)
Now we compute:
\[ \sqrt{2}(y_0 - x_0) = \sqrt{2}\left(-\frac{3\sqrt{2}}{2} - 3\sqrt{2}\right) = \sqrt{2}\left(-\frac{3\sqrt{2}}{2} - \frac{6\sqrt{2}}{2}\right) = \sqrt{2}\left(-\frac{9\sqrt{2}}{2}\right) = -\frac{9}{2} \]

### Final Answer
Thus, the value of \( \sqrt{2}(y_0 - x_0) \) is:
\[ \boxed{-\frac{9}{2}} \]