Find the points on the curve x² + y² – 2x – 3= 0 at whichthe tangents are parallel to x-axis.

Solution

Let P ( x, y ) be any point on the given curve x² + y² - 2 x - 3 = 0.

Tangent to the curve at the point (x, y ) is given by $\frac{dy}{dx}$ .

Differentiating the equation of the cueve w.r.t. x we get

$2 x + 2 y \frac{dy}{dx} - 2 = 0 \Rightarrow \frac{dy}{dx} = \frac{2 - 2 x}{2 y} = \frac{1 - x}{y}$

Let P ( x₁, y₁ ) be the point on the given curve at which the tangents are parallel to the x-axis.

$∴ {\frac{dy}{dx}}_{(x_{1}, y_{1})} = 0 \Rightarrow \frac{1 - x_{1}}{y_{1}} = 0 \Rightarrow 1 - x_{1} = 0 \Rightarrow x_{1} = 1$

To get the value of y₁ just substitute x₁ = 1 in the equation x² + y² - 2 x - 3 = 0, we get

( 1 )² + ( y₁ )² - 2 x 1 - 3 = 0

$\Rightarrow {(y_{1})}^{2} - 4 = 0 \Rightarrow {(y_{1})}^{2} = 4 \Rightarrow y_{1} = \pm 2$

So, the points on the given curve at which the tangents are parallel to the x-axis are ( 1, 2 ) and ( 1, - 2 ).

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