if it is a tangent then the pt must lie both on the line and parabola.

7x+6y=13

x=[13-6y]/7---------------------eq 1

using it in the eq of parabola,we get

y^2 -7[13-6y]/7-8y+14=0

y^2-13+6y-8y+14=0

y^2-2y+1=0

[y-1][y-1]=0  

y=1----------------------------------2

using this in eq1 we get,

x=[13-6]/7

x=7/7

x=1-----------------------------------3

hence the pt of contact is [1,1]

hope u got it!

P.S

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