Question:
Prove that the curves y2 = 4x and x2 + y2 – 6x + 1 = 0 touch each other at the point (1, 2).
Solution:
Given curve equations are: y2 = 4x …. (1) and x2 + y2 – 6x + 1 = 0 ….. (i)
Now, differentiating (i) w.r.t. x, we get
2y.(dy/dx) = 4 ⇒ dy/dx = 2/y
Slope of tangent at (1, 2), m1 = 2/2 = 1
Differentiating (ii) w.r.t. x, we get
2x + 2y.(dy/dx) – 6 = 0
2y. dy/dx = 6 – 2x ⇒ dy/dx = (6 – 2x)/ 2y
Hence, the slope of the tangent at the same point (1, 2)
⇒ m2 = (6 – 2 x 1)/ (2 x 2) = 4/4 = 1
It’s seen that m1 = m2 = 1 at the point (1, 2).
Therefore, the given circles touch each other at the same point (1, 2).