Question:
If the coordinate axes are rotated about the origin through an angle
\[
\frac{\pi}{8}
\]
in the positive direction to remove the \(xy\)-term from the equation
\[
ax^2+bxy+y^2=0,
\]
then
If the coordinate axes are rotated about the origin through an angle
\[
\frac{\pi}{8}
\]
in the positive direction to remove the \(xy\)-term from the equation
\[
ax^2+bxy+y^2=0,
\]
then
Show Hint
For a quadratic equation
\[
Ax^2+Bxy+Cy^2=0,
\]
the angle required to eliminate the \(xy\)-term is determined by
\[
\tan 2\theta=\frac{B}{A-C}.
\]
Substitute the given rotation angle directly into this formula.
Updated On: Jul 29, 2026
- \(a^2+b^2=1\)
- \(a=b+1\)
- \(b=a+1\)
- \(2a=b+5\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For the general second-degree equation
\[
Ax^2+Bxy+Cy^2+\cdots=0,
\]
the angle \(\theta\) through which the axes must be rotated to eliminate the \(xy\)-term is given by
\[
\tan 2\theta=\frac{B}{A-C}.
\]
Step 1: Identify the coefficients. Given equation \[ ax^2+bxy+y^2=0. \] Comparing with \[ Ax^2+Bxy+Cy^2=0, \] we have \[ A=a, \qquad B=b, \qquad C=1. \]
Step 2: Use the condition for removal of the \(xy\)-term. The axes are rotated through \[ \theta=\frac{\pi}{8}. \] Therefore, \[ 2\theta=\frac{\pi}{4}. \] Using \[ \tan 2\theta=\frac{B}{A-C}, \] we get \[ \tan\frac{\pi}{4} = \frac{b}{a-1}. \] \[ 1=\frac{b}{a-1}. \] \[ a-1=b. \] \[ a=b+1. \]
Step 3: Write the final answer. \[ \boxed{a=b+1} \]
Step 1: Identify the coefficients. Given equation \[ ax^2+bxy+y^2=0. \] Comparing with \[ Ax^2+Bxy+Cy^2=0, \] we have \[ A=a, \qquad B=b, \qquad C=1. \]
Step 2: Use the condition for removal of the \(xy\)-term. The axes are rotated through \[ \theta=\frac{\pi}{8}. \] Therefore, \[ 2\theta=\frac{\pi}{4}. \] Using \[ \tan 2\theta=\frac{B}{A-C}, \] we get \[ \tan\frac{\pi}{4} = \frac{b}{a-1}. \] \[ 1=\frac{b}{a-1}. \] \[ a-1=b. \] \[ a=b+1. \]
Step 3: Write the final answer. \[ \boxed{a=b+1} \]
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