Question

The transformed equation of \( 3x^2 - 4xy = r^2 \) when the coordinate axes are rotated about the origin through an angle of \( \tan^{-1}(2) \) in positive direction is

• A. \( x^2 - 4y^2 = r^2 \)
• B. \( 2xy + r^2 = 0 \)
• C. \( 4y^2 - x^2 = r^2 \)
• D. \( xy = r^2 \)

Hint:

Rotation of axes formulas: \( x = X\cos\theta - Y\sin\theta \), \( y = X\sin\theta + Y\cos\theta \). If \( \tan\theta = m \), determine \( \sin\theta \) and \( \cos\theta \) from a right triangle. The angle \( \theta \) that removes the \(xy\) term from \( Ax^2+2Hxy+By^2+\dots=0 \) satisfies \( \tan 2\theta = \frac{2H}{A-B} \). Here, \( \tan\theta = 2 \implies \tan 2\theta = \frac{2(2)}{1-2^2} = \frac{4}{-3} \). For the given equation \(3x^2-4xy=r^2\), \(A=3, 2H=-4, B=0\). So \( \frac{2H}{A-B} = \frac{-4}{3-0} = -4/3 \). This confirms the angle eliminates the \(XY\) term.

Answer 1

The Correct Option is C

Let the angle of rotation be \( \theta \).
Given \( \tan\theta = 2 \).
This implies \( \sin\theta = \frac{2}{\sqrt{5}} \) and \( \cos\theta = \frac{1}{\sqrt{5}} \).
The transformation equations are: \[ x = X\cos\theta - Y\sin\theta = X\left(\frac{1}{\sqrt{5}}\right) - Y\left(\frac{2}{\sqrt{5}}\right) = \frac{X-2Y}{\sqrt{5}} \] \[ y = X\sin\theta + Y\cos\theta = X\left(\frac{2}{\sqrt{5}}\right) + Y\left(\frac{1}{\sqrt{5}}\right) = \frac{2X+Y}{\sqrt{5}} \] Substitute these into \( 3x^2 - 4xy = r^2 \): \[ x^2 = \frac{(X-2Y)^2}{5} = \frac{X^2-4XY+4Y^2}{5} \] \[ xy = \frac{(X-2Y)(2X+Y)}{5} = \frac{2X^2+XY-4XY-2Y^2}{5} = \frac{2X^2-3XY-2Y^2}{5} \] The equation becomes: \[ 3\left(\frac{X^2-4XY+4Y^2}{5}\right) - 4\left(\frac{2X^2-3XY-2Y^2}{5}\right) = r^2 \] Multiply by 5: \[ 3(X^2-4XY+4Y^2) - 4(2X^2-3XY-2Y^2) = 5r^2 \] \[ (3X^2 - 12XY + 12Y^2) - (8X^2 - 12XY - 8Y^2) = 5r^2 \] \[ 3X^2 - 12XY + 12Y^2 - 8X^2 + 12XY + 8Y^2 = 5r^2 \] \[ (3-8)X^2 + (-12+12)XY + (12+8)Y^2 = 5r^2 \] \[ -5X^2 + 0XY + 20Y^2 = 5r^2 \] \[ -5X^2 + 20Y^2 = 5r^2 \] Divide by 5: \[ -X^2 + 4Y^2 = r^2 \implies 4Y^2 - X^2 = r^2 \] Replacing (X,Y) with (x,y): \( 4y^2 - x^2 = r^2 \).
This matches option (3).