If the axes be turned through an angle `tan^-1 2` (in anticlockwise direction), what does the equatio `4xy-3x^2=a^2` become ?

Options

A. `X^(2)+4Y^(2)=a^(2)`
B. `X^(2)-4Y^(2)=4a^(2)`
C. `X^(2)-4Y^(2)=a^(2)`
D. `X^(2)+4Y^(2)=4a^(2)`

Explanation

To solve the problem, we need to find the new equation after turning the axes through an angle of \( \theta = \tan^{-1}(2) \) in the anticlockwise direction. The original equation is given as \( 4xy - 3x^2 = a^2 \).

### Step-by-Step Solution:

1. **Determine the angle of rotation**:
\[ \theta = \tan^{-1}(2) \]
This means \( \tan(\theta) = 2 \), which implies that the opposite side (perpendicular) is 2 and the adjacent side (base) is 1.

2. **Calculate \( \sin(\theta) \) and \( \cos(\theta) \)**:
Using the right triangle formed, we can calculate the hypotenuse:
\[ \text{Hypotenuse} = \sqrt{2^2 + 1^2} = \sqrt{5} \]
Now, we can find \( \sin(\theta) \) and \( \cos(\theta) \):
\[ \sin(\theta) = \frac{2}{\sqrt{5}}, \quad \cos(\theta) = \frac{1}{\sqrt{5}} \]

3. **Transform the coordinates**:
The new coordinates \( (x', y') \) after rotation are given by:
\[ x' = x \cos(\theta) + y \sin(\theta) = x \cdot \frac{1}{\sqrt{5}} + y \cdot \frac{2}{\sqrt{5}} = \frac{x + 2y}{\sqrt{5}} \]
\[ y' = -x \sin(\theta) + y \cos(\theta) = -x \cdot \frac{2}{\sqrt{5}} + y \cdot \frac{1}{\sqrt{5}} = \frac{-2x + y}{\sqrt{5}} \]

4. **Substitute the new coordinates into the original equation**:
The original equation is:
\[ 4xy - 3x^2 = a^2 \]
Substitute \( x \) and \( y \) with \( x' \) and \( y' \):
\[ 4\left(\frac{x + 2y}{\sqrt{5}}\right)\left(\frac{-2x + y}{\sqrt{5}}\right) - 3\left(\frac{x + 2y}{\sqrt{5}}\right)^2 = a^2 \]

5. **Simplify the equation**:
Expanding the terms:
\[ 4 \cdot \frac{(x + 2y)(-2x + y)}{5} - 3 \cdot \frac{(x + 2y)^2}{5} = a^2 \]
This simplifies to:
\[ \frac{4(-2x^2 + xy + 4y^2)}{5} - \frac{3(x^2 + 4xy + 4y^2)}{5} = a^2 \]
Combine the fractions:
\[ \frac{4(-2x^2 + xy + 4y^2) - 3(x^2 + 4xy + 4y^2)}{5} = a^2 \]

6. **Combine like terms**:
\[ = \frac{-8x^2 + 4xy + 16y^2 - 3x^2 - 12xy - 12y^2}{5} = a^2 \]
Simplifying the numerator:
\[ = \frac{-11x^2 - 8xy + 4y^2}{5} = a^2 \]

7. **Final form**:
Multiply through by 5 to eliminate the denominator:
\[ -11x^2 - 8xy + 4y^2 = 5a^2 \]
Rearranging gives:
\[ 11x^2 + 8xy - 4y^2 = -5a^2 \]

### Final Answer:
The transformed equation after rotating the axes is:
\[ 11x^2 + 8xy - 4y^2 = -5a^2 \]