Question

Which one of the following equations is obtained from the equation \(xy=-8\) by rotating the axes counterclockwise through an angle of \(45^\circ\)? (Here \(X, Y\) denote the new coordinates.)

• A. \(X^2 - Y^2 + 8 = 0\)
• B. \(X^2 - Y^2 + 16 = 0\)
• C. \(X^2 - Y^2 + 24 = 0\)
• D. \(X^2 - Y^2 + 32 = 0\)

Hint:

Substitute \(x=\dfrac{X-Y}{\sqrt2}, y=\dfrac{X+Y}{\sqrt2}\) into \(xy=-8\).

Answer 1

The Correct Option is B

Step 1: The rotation formulas that express old coordinates in terms of new coordinates, when the axes are rotated counterclockwise through angle \(\theta\), are \[ x = X\cos\theta - Y\sin\theta, \qquad y = X\sin\theta + Y\cos\theta \] For \(\theta = 45^\circ\), \(\cos45^\circ=\sin45^\circ=\dfrac{1}{\sqrt2}\), so \[ x = \frac{X-Y}{\sqrt2}, \qquad y = \frac{X+Y}{\sqrt2} \]

Step 2: Substitute into \(xy=-8\): \[ xy = \frac{(X-Y)}{\sqrt2}\cdot\frac{(X+Y)}{\sqrt2} = \frac{X^2-Y^2}{2} \]

Step 3: Set this equal to \(-8\) and simplify: \[ \frac{X^2-Y^2}{2} = -8 \implies X^2-Y^2 = -16 \implies X^2-Y^2+16=0 \] \[ \boxed{X^2-Y^2+16=0} \]