AP EAMCET Engineering 18 Sep 2020 Shift 1 Paper

Let the point of tangency of tangent
x−√3y+9=0 to the parabola y2=12x is (x1y1)
Since equation of tangent to the parabola y2=12x at point (x1,y1) is yy1=6(x+x1)
6x−y1y+6x1=0, which represent the tangent x−√3y+9=0, so on comparing, we get
Therefore, point of contact is (9,6√3).