The tangents to the curve y = (x - 2)^2 - 1 at its points of intersection with the line x - y = 3, intersect at the point.

Correct option is C (52,-1;)
Step 1: Find the points of intersection of given curve and line
Given curve is y=(x-2)^2-1;and equation of line is x-y=3
Solving for x , we have
⇒x-3=(x-2)^2-1;
⇒x-3=x^2+4-4x-1;
⇒x^2-5x+6=0
⇒x^2-3x-2x+6=0[Splitting middle term]
⇒x(x-3)-2(x-3)=0
⇒(x-2)(x-3)=0
⇒x=2,3
At x=2,y=-1; and at x=3,y=0
Hence, points of intersection are : A(2,-1) and B(3,0)
Step 2: Find the equation of tangents at these points
For curve is y=(x-2)^2-1;
Differentiating w.r.t. x on both sides we have
Slope of tangent at any point, dy/dx=2(x-2)
At point A(2,-1) , slope is =0
Hence equation of tangent is y=-1; (1)
At point B(3,0) , slope is =2
Hence equation of tangent is y-0=2(x-3).. (2)[Using one-point slope form of equation of line, y-y1=m(x-x1)]
Point of intersection of tangents 1 and 2 is (52,-1;)
Hence, Option 'C' is correct