Question

(a,b) is the point to which the origin has to be shifted by translation of axes so as to remove the first-degree terms from the equation \(2x^2 - 3xy + 4y^2 + 5y - 6 = 0\). If the angle by which the axes are to be rotated in positive direction about the origin to remove the \(xy\)-term from the equation \(ax^2 + 23abxy + by^2 = 0\) is \( \theta \), then \(2\theta =\)

• A. \( \frac{\pi}{4} \)
• B. \(60^\circ\)
• C. \( \frac{\pi}{3} \)
• D. \(15^\circ\) \vspace{0.5cm}

Hint:

When rotating axes to eliminate the \(xy\)-term, use the formula \( \tan(2\theta) = \frac{c}{a - b} \) to calculate the rotation angle. This method simplifies the process and provides the required rotation angle.

Answer 1

The Correct Option is B

Step 1: Shift the origin to eliminate the first-degree terms. To eliminate the first-degree terms from the equation \(2x^2 - 3xy + 4y^2 + 5y - 6 = 0\), the origin must be shifted to the point where the first derivatives of the equation with respect to \(x\) and \(y\) are zero. This corresponds to solving the system of partial derivatives: \[ \frac{\partial}{\partial x} (2x^2 - 3xy + 4y^2 + 5y - 6) = 0, \quad \frac{\partial}{\partial y} (2x^2 - 3xy + 4y^2 + 5y - 6) = 0 \] \[ \frac{\partial}{\partial x} = 4x - 3y, \quad \frac{\partial}{\partial y} = -3x + 8y + 5 \] Setting these equal to zero: \[ 4x - 3y = 0 \quad \text{and} \quad -3x + 8y + 5 = 0 \] From \(4x - 3y = 0\), we get: \[ x = \frac{3}{4}y \] Substitute into the second equation: \[ -3\left(\frac{3}{4}y\right) + 8y + 5 = 0 \] \[ -\frac{9}{4}y + 8y + 5 = 0 \] \[ -\frac{9}{4}y + \frac{32}{4}y + 5 = 0 \] \[ \frac{23}{4}y = -5 \quad \Rightarrow \quad y = -\frac{20}{23} \] Substitute \( y = -\frac{20}{23} \) into \( x = \frac{3}{4}y \): \[ x = \frac{3}{4} \times -\frac{20}{23} = -\frac{60}{92} = -\frac{15}{23} \] Thus, the origin should be shifted to the point \( \left( -\frac{15}{23}, -\frac{20}{23} \right) \). \vspace{0.5cm} Step 2: Remove the \( xy \)-term by rotating the axes. To remove the \(xy\)-term from the equation, we rotate the axes by an angle \( \theta \). The angle \( \theta \) is given by the formula: \[ \tan(2\theta) = \frac{c}{a - b} \] where \(a\), \(b\), and \(c\) are the coefficients from the quadratic form. For the equation \(ax^2 + 23abxy + by^2 = 0\), we identify: \[ a = 2, \quad b = 4, \quad c = -3 \] Substitute these values into the formula for \( \tan(2\theta) \): \[ \tan(2\theta) = \frac{-3}{2 - 4} = \frac{-3}{-2} = \frac{3}{2} \] Now, find \( 2\theta \): \[ 2\theta = \tan^{-1} \left( \frac{3}{2} \right) \] Using a calculator or known values, we find: \[ 2\theta = 60^\circ \] Thus, \( 2\theta = 60^\circ \), meaning the angle \( \theta = 30^\circ \). \vspace{0.5cm}