if the axes are rotated through 60 in the anticlockwise sense,find the transformed form of the equation `x^2-y^2=a^2`,

Options

A. ` X^(2) + Y^(2) - 3 sqrt(3) XY = 2a^(2) `
B. `X^(2) + Y^(2) = a^(2)`
C. ` Y^(2) - X^(2) - 2 sqrt(3) XY = 2a^(2) `
D. `X^(2) - Y^(2) + 2 sqrt(3) XY = 2a^(2) `

Answer

To solve the problem of finding the transformed form of the equation \( x^2 - y^2 = a^2 \) when the axes are rotated through 60 degrees in the anticlockwise direction, we will follow these steps:

### Step 1: Understand the Rotation of Axes
When the axes are rotated by an angle \( \theta \), the new coordinates \( (X, Y) \) can be expressed in terms of the old coordinates \( (x, y) \) using the following transformation equations:
\[ X = x \cos \theta - y \sin \theta \]
\[ Y = x \sin \theta + y \cos \theta \]
For \( \theta = 60^\circ \):
\[ \cos 60^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2} \]

### Step 2: Substitute the Values
Substituting \( \cos 60^\circ \) and \( \sin 60^\circ \) into the transformation equations gives:
\[ X = x \cdot \frac{1}{2} - y \cdot \frac{\sqrt{3}}{2} \]
\[ Y = x \cdot \frac{\sqrt{3}}{2} + y \cdot \frac{1}{2} \]

### Step 3: Solve for \( x \) and \( y \)
We need to express \( x \) and \( y \) in terms of \( X \) and \( Y \). Rearranging the equations:
1. From the first equation:
\[ x = 2X + \sqrt{3}Y \]
2. From the second equation:
\[ y = \sqrt{3}X - 2Y \]

### Step 4: Substitute into the Original Equation
Now substitute \( x \) and \( y \) into the original equation \( x^2 - y^2 = a^2 \):
\[ (2X + \sqrt{3}Y)^2 - (\sqrt{3}X - 2Y)^2 = a^2 \]

### Step 5: Expand the Squares
Expanding both squares:
1. \( (2X + \sqrt{3}Y)^2 = 4X^2 + 4\sqrt{3}XY + 3Y^2 \)
2. \( (\sqrt{3}X - 2Y)^2 = 3X^2 - 4\sqrt{3}XY + 4Y^2 \)

### Step 6: Combine the Results
Now, substituting back into the equation:
\[ (4X^2 + 4\sqrt{3}XY + 3Y^2) - (3X^2 - 4\sqrt{3}XY + 4Y^2) = a^2 \]
This simplifies to:
\[ 4X^2 + 4\sqrt{3}XY + 3Y^2 - 3X^2 + 4\sqrt{3}XY - 4Y^2 = a^2 \]
Combining like terms:
\[ (4X^2 - 3X^2) + (4\sqrt{3}XY + 4\sqrt{3}XY) + (3Y^2 - 4Y^2) = a^2 \]
This results in:
\[ X^2 + 8\sqrt{3}XY - Y^2 = a^2 \]

### Step 7: Final Form
Rearranging gives us the final transformed equation:
\[ X^2 - Y^2 + 8\sqrt{3}XY - a^2 = 0 \]