Answer :

Given:


Curves y2 = 4x ...(1)


& x2 + y2 – 6x + 1 = 0 ...(2)


The point of intersection of two curves is (1,2)


First curve is y2 = 4x


Differentiating above w.r.t x,


2y. = 4


y. = 2


m1 ...(3)


Second curve is x2 + y2 – 6x + 1 = 0


2x + 2y.6 – 0 = 0


x + y.3 = 0


y. = 3 – x


...(4)


At (1,2),we have,


m1



m1 = 1


At (1,2),we have,


m2



m2 = 1


Clearly, m1 = m2 = 1 at (1,2)


So, given curve touch each other at (1,2)


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