Find the point on the curve `y^2 - x^2 + 2x - 1 =0`
where the tangent is parallel to the x - axis.

Explanation

Given the curve is `y^2 - x^2 + 2xx - 1 = 0…(1)`
`rArr 2y dy/dx - 2xx + 2 = 0` `rArr dy/dx = x- 1/y`
if the tangent is parallel to
`x-axis then dy/dx = 0`
`x - 1/y = 0 rArr x = 1`
Putting x = 1 in (1) we get
`y^2 - 1 + 2 - 1 = 0 rArr y = 0`
The point is (1,0).