To find the point where the tangent to the curve [tex]\( y^2 - x^2 + 2x - 1 = 0 \)[/tex] is parallel to the x-axis, we need to determine when the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] ([tex]\( \frac{dy}{dx} \)[/tex]) is equal to zero. This denotes a horizontal tangent, which is parallel to the x-axis.
Let's follow these steps to solve the problem:
Step 1: Implicit Differentiation
Given the curve equation:
[tex]\[ y^2 - x^2 + 2x - 1 = 0 \][/tex]
Implicitly differentiate both sides of the equation with respect to [tex]\( x \)[/tex]:
[tex]\[ \frac{d}{dx} (y^2 - x^2 + 2x - 1) = 0 \][/tex]
By applying the differentiation rules:
[tex]\[ 2y \frac{dy}{dx} - 2x + 2 = 0 \][/tex]
Step 2: Solve for [tex]\(\frac{dy}{dx}\)[/tex]
Rearrange the terms to solve for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ 2y \frac{dy}{dx} = 2x - 2 \][/tex]
Divide by [tex]\( 2y \)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{2x - 2}{2y} = \frac{x - 1}{y} \][/tex]
Step 3: Set the Derivative Equal to Zero
For the tangent to be parallel to the x-axis, [tex]\(\frac{dy}{dx} = 0\)[/tex]:
[tex]\[ \frac{x - 1}{y} = 0 \][/tex]
Solve for [tex]\( x \)[/tex]:
[tex]\[ x - 1 = 0 \][/tex]
Therefore,
[tex]\[ x = 1 \][/tex]
Step 4: Find Corresponding [tex]\( y \)[/tex] Values
Substitute [tex]\( x = 1 \)[/tex] back into the original curve equation to find the corresponding [tex]\( y \)[/tex] values:
[tex]\[ y^2 - (1)^2 + 2(1) - 1 = 0 \][/tex]
Simplify:
[tex]\[ y^2 - 1 + 2 - 1 = 0 \][/tex]
[tex]\[ y^2 = 0 \][/tex]
[tex]\[ y = 0 \][/tex]
Step 5: Determine the Point
The point where the tangent to the curve [tex]\( y^2 - x^2 + 2x - 1 = 0 \)[/tex] is parallel to the x-axis is:
[tex]\[ (x, y) = (1, 0) \][/tex]
Therefore, the point where the tangent to the curve is parallel to the x-axis is [tex]\(\boxed{(1, 0)}\)[/tex].
