y^2 − 4y(−2y) = 4y^2 + 5
5 y^2 = 5.Thus y = ±1 and x = ±2. The points, then, are (2, −1) and (−2, 1).
Example 8 __
Find the x-coordinate of any point on the curve of y = sin^2 (x + 1) for which the
tangent is parallel to the line 3x − 3y − 5 = 0.
SOLUTION: Since = 2sin(x + 1) cos(x + 1) = sin2(x + 1) and since the given
line has slope 1, we seek x such that sin 2(x + 1) = 1. Then
2(x + 1) = + 2 nπ (n an integer)