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 y – axis.
Answer» Since, the tangent is parallel to y – axis, its slope is not defined, then the normal is parallel to x – axis whose slope is zero. i.e, \(\cfrac{-1}{\frac{dy}{dx}}\) = 0 ⇒ \(\cfrac{-1}{\frac{1-x}{y}}\) = 0 ⇒ \(\cfrac{-y}{1-x}\) = 0 ⇒ y = 0 Substituting y = 0 in x² + y² – 2x – 3 = 0, ⇒ x² + 0² – 2×x – 3 = 0 ⇒ x² – 2x – 3 = 0 Using factorization method, we can solve above quadratic equation ⇒ x²– 3x + x – 3 = 0 ⇒ x(x – 3) + 1(x – 3) = 0 ⇒ (x – 3)(x + 1) = 0 ⇒ x = 3 & x = – 1 Thus, the required point is (3,0) & ( – 1,0)

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

Answer»

Since, the tangent is parallel to y – axis, its slope is not defined, then the normal is parallel to x – axis whose slope is zero.

i.e, \(\cfrac{-1}{\frac{dy}{dx}}\) = 0

⇒ \(\cfrac{-1}{\frac{1-x}{y}}\) = 0

⇒ \(\cfrac{-y}{1-x}\) = 0

⇒ y = 0

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

⇒ x² + 0² – 2×x – 3 = 0

⇒ x² – 2x – 3 = 0

Using factorization method, we can solve above quadratic equation

⇒ x²– 3x + x – 3 = 0

⇒ x(x – 3) + 1(x – 3) = 0

⇒ (x – 3)(x + 1) = 0

⇒ x = 3 & x = – 1

Thus, the required point is (3,0) & ( – 1,0)