Final answer:
The points on the curve x²+y²−2x−3=0 where the tangents are parallel to the x-axis are found by computing where the implicit derivative equals zero, indicating a horizontal tangent. Completing these calculations, the points are (3, 0) and (-1, 0).
Step-by-step explanation:
To find the points on the curve x²+y²−2x−3=0 where the tangents are parallel to the x-axis, we need to calculate the derivative of the function to find the slope of the tangent. A tangent parallel to the x-axis has a slope of 0, i.e., the derivative at that point is zero. The given curve is a circle with its center shifted from the origin. By completing the square, we express the equation in the standard form of a circle's equation.
First, we rewrite the equation as:
(x² - 2x + 1) + (y²) = 1 + 3
(x - 1)² + y² = 4
The circle has a center at (1, 0) and a radius of 2. To find where the derivatives are zero, we differentiate the equation implicitly w.r.t. x:
2x - 2 + 2yy' = 0
y' (the derivative of y w.r.t. x) represents the slope of the tangent, so we set y' = 0 for a horizontal tangent:
2yy' = 0 → y = 0
Since we know y = 0 for horizontal tangents, plug it into the original equation:
x² - 2x - 3 = 0
The solutions give us the x-coordinates of the points where the tangents are horizontal. Completing the computation for x:
(x - 3)(x + 1) = 0 → x = 3 or x = -1
Thus, the points on the curve where the tangents are parallel to the x-axis are (3, 0) and (-1, 0).