Find the point on the curve x2 + y2 – 2x – 3 = 0 at which the tang

1.	Find the point on the curve x2 + y2 – 2x – 3 = 0 at which the tangents are parallel to the x – axis.
Answer» Given: The curve is x² + y² – 2x – 3 = 0 Differentiating the above w.r.t x, we get The Slope of tangent, ⇒ 2x^{2 – 1} + 2y^{2 – 1}\(\frac{dy}{dx}\)– 2 – 0 = 0 ⇒ 2x + 2y \(\frac{dy}{dx}\) – 2 = 0 ⇒ 2y \(\frac{dy}{dx}\) = 2 – 2x ⇒ \(\frac{dy}{dx}\) = \(\frac{2-2x}{2y}\) ⇒\(\frac{dy}{dx}\) = \(\frac{1-x}{y}\)...(1) (i) Since, the tangent is parallel to x – axis ⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2) \(\therefore\) tan(0) = 0 \(\therefore\frac{dy}{dx}\) = The Slope of the tangent = tanθ From (1) & (2),we get, ⇒ \(\frac{1-x}{y}\) = 0 ⇒ 1 – x = 0 ⇒ x = 1 Substituting x = 1 in x² + y² – 2x – 3 = 0, ⇒ 1² + y² – 2×1 – 3 = 0 ⇒ y² – 4 = 0 ⇒ y² = 4 ⇒ y = ± 2 Thus, the required point is (1,2) & (1, – 2)

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

Answer»

Given:

The curve is x² + y² – 2x – 3 = 0

Differentiating the above w.r.t x, we get The Slope of tangent,

⇒ 2x^{2 – 1} + 2y^{2 – 1}\(\frac{dy}{dx}\)– 2 – 0 = 0

⇒ 2x + 2y \(\frac{dy}{dx}\) – 2 = 0

⇒ 2y \(\frac{dy}{dx}\) = 2 – 2x

⇒ \(\frac{dy}{dx}\) = \(\frac{2-2x}{2y}\)

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

(i) Since, the tangent is parallel to x – axis

⇒ \(\frac{dy}{dx}\) = tan(0) = 0 ...(2)

\(\therefore\) tan(0) = 0

\(\therefore\frac{dy}{dx}\) = The Slope of the tangent = tanθ

From (1) & (2),we get,

⇒ \(\frac{1-x}{y}\) = 0

⇒ 1 – x = 0

⇒ x = 1

Substituting x = 1 in x² + y² – 2x – 3 = 0,

⇒ 1² + y² – 2×1 – 3 = 0

⇒ y² – 4 = 0

⇒ y² = 4

⇒ y = ± 2

Thus, the required point is (1,2) & (1, – 2)