Question:
If the axes are rotated about the origin in positive direction through an angle of \(30^\circ\), then the transformed equation of
\[
x^2+2\sqrt{3}xy-y^2=2a^2
\]
is
If the axes are rotated about the origin in positive direction through an angle of \(30^\circ\), then the transformed equation of
\[
x^2+2\sqrt{3}xy-y^2=2a^2
\]
is
Show Hint
For rotation of axes problems, first compute
\[
\tan2\theta=\frac{2H}{A-B}.
\]
A suitable rotation removes the \(xy\)-term and simplifies the equation into standard form.
Updated On: Jun 17, 2026
- \(x'^2+y'^2=a^2\)
- \(x'^2+y'^2=2a^2\)
- \(x'^2-y'^2=a^2\)
- \(x'^2-y'^2=2a^2\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
When axes are rotated through an angle \(\theta\), the new coordinates are related to the old coordinates by:
\[
x=x'\cos\theta-y'\sin\theta,
\]
\[
y=x'\sin\theta+y'\cos\theta.
\]
The purpose of rotation is usually to eliminate the \(xy\)-term from the equation.
Step 1: Identify coefficients of the quadratic equation.
Given equation: \[ x^2+2\sqrt3xy-y^2=2a^2. \] Comparing with the standard form \[ Ax^2+2Hxy+By^2=0, \] we obtain \[ A=1,\qquad H=\sqrt3,\qquad B=-1. \]
Step 2: Find angle of rotation.
The angle \(\theta\) required to eliminate the \(xy\)-term is determined by \[ \tan2\theta=\frac{2H}{A-B}. \] Substituting values, \[ \tan2\theta=\frac{2\sqrt3}{1-(-1)} =\frac{2\sqrt3}{2} =\sqrt3. \] Therefore, \[ 2\theta=60^\circ \] and hence \[ \theta=30^\circ. \] This matches the given rotation.
Step 3: Find transformed coefficients.
The transformed coefficients are: \[ A'=A\cos^2\theta+2H\sin\theta\cos\theta+B\sin^2\theta, \] \[ B'=A\sin^2\theta-2H\sin\theta\cos\theta+B\cos^2\theta. \] Using \[ \cos30^\circ=\frac{\sqrt3}{2}, \qquad \sin30^\circ=\frac12, \] we get \[ A' =1\cdot\frac34 +2\sqrt3\cdot\frac{\sqrt3}{4} -1\cdot\frac14. \] Thus, \[ A'=\frac34+\frac{6}{4}-\frac14 =\frac{8}{4} =2. \] Similarly, \[ B' =1\cdot\frac14 -2\sqrt3\cdot\frac{\sqrt3}{4} -1\cdot\frac34. \] Hence, \[ B'=\frac14-\frac64-\frac34 =-2. \] Therefore transformed equation becomes \[ 2x'^2-2y'^2=2a^2. \] Dividing throughout by \(2\), \[ x'^2-y'^2=a^2. \] Hence the required transformed equation is \[ \boxed{x'^2-y'^2=a^2}. \]
Step 1: Identify coefficients of the quadratic equation.
Given equation: \[ x^2+2\sqrt3xy-y^2=2a^2. \] Comparing with the standard form \[ Ax^2+2Hxy+By^2=0, \] we obtain \[ A=1,\qquad H=\sqrt3,\qquad B=-1. \]
Step 2: Find angle of rotation.
The angle \(\theta\) required to eliminate the \(xy\)-term is determined by \[ \tan2\theta=\frac{2H}{A-B}. \] Substituting values, \[ \tan2\theta=\frac{2\sqrt3}{1-(-1)} =\frac{2\sqrt3}{2} =\sqrt3. \] Therefore, \[ 2\theta=60^\circ \] and hence \[ \theta=30^\circ. \] This matches the given rotation.
Step 3: Find transformed coefficients.
The transformed coefficients are: \[ A'=A\cos^2\theta+2H\sin\theta\cos\theta+B\sin^2\theta, \] \[ B'=A\sin^2\theta-2H\sin\theta\cos\theta+B\cos^2\theta. \] Using \[ \cos30^\circ=\frac{\sqrt3}{2}, \qquad \sin30^\circ=\frac12, \] we get \[ A' =1\cdot\frac34 +2\sqrt3\cdot\frac{\sqrt3}{4} -1\cdot\frac14. \] Thus, \[ A'=\frac34+\frac{6}{4}-\frac14 =\frac{8}{4} =2. \] Similarly, \[ B' =1\cdot\frac14 -2\sqrt3\cdot\frac{\sqrt3}{4} -1\cdot\frac34. \] Hence, \[ B'=\frac14-\frac64-\frac34 =-2. \] Therefore transformed equation becomes \[ 2x'^2-2y'^2=2a^2. \] Dividing throughout by \(2\), \[ x'^2-y'^2=a^2. \] Hence the required transformed equation is \[ \boxed{x'^2-y'^2=a^2}. \]
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