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

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, = 0


= 0


= 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)


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