Question:
The origin is shifted to (2, 3) and axes are rotated through angle \(\theta\). If \(3x^2+2xy+3y^2-18x-22y+50=0\) transforms to \(4x^2+2y^2-1=0\), then \(\theta =\)
The origin is shifted to (2, 3) and axes are rotated through angle \(\theta\). If \(3x^2+2xy+3y^2-18x-22y+50=0\) transforms to \(4x^2+2y^2-1=0\), then \(\theta =\)
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If \(A=C\) in \(Ax^2+Bxy+Cy^2\), then immediately \(\theta=\frac{\pi}{4}\).
Updated On: Jun 9, 2026
- \(\frac{\pi}{4}\)
- \(\frac{\pi}{6}\)
- \(\frac{\pi}{3}\)
- \(\frac{\pi}{2}\)
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The Correct Option is A
Solution and Explanation
Concept:
For a second-degree equation
\[
Ax^2+Bxy+Cy^2+Dx+Ey+F=0,
\]
the angle of rotation required to eliminate the \(xy\)-term is given by
\[
\tan 2\theta=\frac{B}{A-C}.
\]
Step 1: Identify the coefficients of the quadratic terms. Given \[ 3x^2+2xy+3y^2-18x-22y+50=0 \] Comparing with \[ Ax^2+Bxy+Cy^2+Dx+Ey+F=0, \] we obtain \[ A=3,\qquad B=2,\qquad C=3. \]
Step 2: Apply the rotation formula. Using \[ \tan 2\theta = \frac{B}{A-C} \] gives \[ \tan 2\theta = \frac{2}{3-3} = \infty \] Hence \[ 2\theta=\frac{\pi}{2} \]
Step 3: Find \(\theta\). \[ \theta=\frac{\pi}{4} \] Therefore, \[ \boxed{\theta=\frac{\pi}{4}} \]
Step 1: Identify the coefficients of the quadratic terms. Given \[ 3x^2+2xy+3y^2-18x-22y+50=0 \] Comparing with \[ Ax^2+Bxy+Cy^2+Dx+Ey+F=0, \] we obtain \[ A=3,\qquad B=2,\qquad C=3. \]
Step 2: Apply the rotation formula. Using \[ \tan 2\theta = \frac{B}{A-C} \] gives \[ \tan 2\theta = \frac{2}{3-3} = \infty \] Hence \[ 2\theta=\frac{\pi}{2} \]
Step 3: Find \(\theta\). \[ \theta=\frac{\pi}{4} \] Therefore, \[ \boxed{\theta=\frac{\pi}{4}} \]
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