Shifting or origin (0,0) to (b,k)
`f(x,y)impliesf(x+h,y+k)`
Rotation of axes through an angle `theta`
`f(x,y)impliesf(x cos theta-y sin theta,x sin theta+ycos theta)`
Now Answer the following questions:
Axes are rotated through a+ive obtuse angle `theta` so that the transformed equation of the curve `3x^(2)-6xy+3y^(2)+7x-3=0` is free from the term of xy then the coefficient of `x^(2)` in the transformed equation is...............

Answer

To solve the problem, we need to transform the given equation \(3x^2 - 6xy + 3y^2 + 7x - 3 = 0\) by rotating the axes through an obtuse angle \(\theta\) so that the transformed equation is free from the \(xy\) term. We will follow these steps:

### Step 1: Identify the transformation equations
When we rotate the axes through an angle \(\theta\), the new coordinates \((x', y')\) are given by:
\[ x' = x \cos \theta - y \sin \theta \]
\[ y' = x \sin \theta + y \cos \theta \]

### Step 2: Substitute the transformation into the original equation
We substitute \(x\) and \(y\) in terms of \(x'\) and \(y'\):
\[ x = x' \cos \theta + y' \sin \theta \]
\[ y = -x' \sin \theta + y' \cos \theta \]

### Step 3: Substitute into the original equation
We substitute these expressions into the original equation:
\[ 3(x' \cos \theta + y' \sin \theta)^2 - 6(x' \cos \theta + y' \sin \theta)(-x' \sin \theta + y' \cos \theta) + 3(-x' \sin \theta + y' \cos \theta)^2 + 7(x' \cos \theta + y' \sin \theta) - 3 = 0 \]

### Step 4: Expand the equation
We need to expand each term carefully:
1. Expand \(3(x' \cos \theta + y' \sin \theta)^2\)
2. Expand \(-6(x' \cos \theta + y' \sin \theta)(-x' \sin \theta + y' \cos \theta)\)
3. Expand \(3(-x' \sin \theta + y' \cos \theta)^2\)
4. Expand \(7(x' \cos \theta + y' \sin \theta)\)

### Step 5: Collect like terms
After expanding, we will collect the coefficients of \(x'^2\), \(y'^2\), and \(x'y'\).

### Step 6: Set the coefficient of \(x'y'\) to zero
To eliminate the \(x'y'\) term, we set the coefficient of \(x'y'\) to zero. The coefficient of \(x'y'\) will be:
\[ -6 \sin \theta \cos \theta + 3 \sin 2\theta = 0 \]
This simplifies to:
\[ -6 \sin \theta \cos \theta + 3 \sin 2\theta = 0 \]

### Step 7: Solve for \(\theta\)
From the equation above, we can solve for \(\theta\). Since we are looking for a positive obtuse angle, we can find that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
This leads to:
\[ -6 + 3 = 0 \]
Thus, we can find \(\theta\) as \(135^\circ\).

### Step 8: Find the coefficient of \(x'^2\)
The coefficient of \(x'^2\) in the transformed equation is given by:
\[ 3 \cos^2 \theta - 6 \sin \theta \cos \theta + 3 \sin^2 \theta \]
Substituting \(\theta = 135^\circ\):
- \(\cos 135^\circ = -\frac{1}{\sqrt{2}}\)
- \(\sin 135^\circ = \frac{1}{\sqrt{2}}\)

Calculating:
\[ 3 \left(-\frac{1}{\sqrt{2}}\right)^2 - 6 \left(-\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right) + 3 \left(\frac{1}{\sqrt{2}}\right)^2 \]
This simplifies to:
\[ 3 \cdot \frac{1}{2} + 6 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2} = \frac{3}{2} + 3 + \frac{3}{2} = 6 \]

### Final Answer
Thus, the coefficient of \(x'^2\) in the transformed equation is \(6\).