Question

Find the points on the curve \(x^2 + y^2 − 2x − 3 = 0\) at which the tangents are parallel to the x-axis.

Answer 1

The equation of the given curve is x² + y² − 2x − 3 = 0.

On differentiating with respect to x, we have:

2x + 2y \(\frac {dy}{dx}\) - 2 = 0

y\(\frac {dy}{dx}\) = 1 - x

\(\frac {dy}{dx}\) = \(\frac {1-x}{y}\)

Now, the tangents are parallel to the x-axis if the slope of the tangent is 0.

\(\frac {1-x}{y}\)= 0 \(\implies\)1-x = 0 \(\implies\)x = 1

But, x² + y² − 2x − 3 = 0 for x = 1. 

y² = 4

\(\implies\)y²=4\(\implies\)y=±2

∴ Hence, the points at which the tangents are parallel to the x-axis are (1, 2) and (1, −2).