91Ó°ÊÓ

Replace x with \(r \cos\theta\) and y with \(r\sin\theta\) in the equation: \((r\cos\theta)^2 - (r\sin\theta)^2 = 1\).

By squaring both terms, we have: \(r^2\cos^2\theta - r^2\sin^2\theta = 1\) Factor out the common term, \(r^2\), yields: \(r^2(\cos^2\theta - \sin^2\theta) = 1\).

Divide both sides of the equation by \((\cos^2\theta - \sin^2\theta)\) to isolate r: \(r^2 = \frac{1}{\cos^2\theta - \sin^2\theta}\) Then take the square root of both sides in order to obtain r: \(r = \sqrt{\frac{1}{\cos^2\theta - \sin^2\theta}}\).

The given polar equation, \(r = \sqrt{\frac{1}{\cos^2\theta - \sin^2\theta}}\), represents a hyperbola in polar coordinates. The numerator of the fraction is constant 1, and the denominator is a difference of squares.

To sketch the curve, we will plot a few points for various values of θ and remember that r is dependent on θ: 1. For θ = 0, the denominator becomes 1, so the point at (1,0) will be plotted. 2. For θ = π/4, the denominator also becomes 1, so we plot another point at \((\sqrt{2}/2, \sqrt{2}/2)\). 3. For θ = π/2, the denominator becomes -1, which is undefined. This indicates that there will be a branch of the hyperbola in the direction of θ = π/2. 4. Similar to the case for θ = π/2, there will be branches of the hyperbola in the directions θ = 3π/2, θ = π, and θ = 2π, as the denominator becomes undefined in these cases. 5. For θ between 0 and π/2, the denominator is positive, so we will have points in the first quadrant. 6. For θ between π/2 and π, the denominator is negative, so there will be no points in the second quadrant. 7. For θ between π and 3π/2, the denominator is positive, so we will have points in the third quadrant. 8. For θ between 3π/2 and 2π, the denominator is negative, so we will have no points in the fourth quadrant. Finally, connect the points and form the branches of the hyperbola. The curve will have two branches in the first and third quadrants, with their branches pointing towards θ = π/2, 3π/2, π, and 2π.