# Find the points on the curve x^{2} + y^{2} - 2x - 3 = 0 at which the tangents are parallel to the x-axis

**Solution:**

For a curve y = f(x) containing the point (x_{1},y_{1}) the equation of the tangent line to the curve at (x_{1},y_{1}) is given by

y − y_{1} = f′(x_{1}) (x − x_{1})

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

The equation of the given curve is x^{2} + y^{2} - 2x - 3 = 0

On differentiating with respect to x , we have:

2x + 2y dy/dx - 2 = 0

⇒ y dy/dx

= 1- x

⇒ dy/dx

= (1 - x)/y

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

Therefore,

(1 - x)/y = 0

⇒ 1- x = 0

⇒ x = 1

But we have x^{2} + y^{2} - 2x - 3 = 0 for x = 1

Hence,

⇒ y^{2} = 4

⇒ y = ± 2

Thus, the points are (1, 2) and (1, - 2)

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 19

## Find the points on the curve x^{2} + y^{2} - 2x - 3 = 0 at which the tangents are parallel to the x-axis.

**Summary:**

The points on the curve x^{2} + y^{2} - 2x - 3 = 0 at which the tangents are parallel to the x-axis are (1, 2) and (1, - 2)

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