1: Step 1: Let the tangent equation be Y=kx+b (solving the tangent equation is to determine the values of k and b).

Step 2: Because the tangent passes through point (5.3), there are 5=3k+b, and both are b=5-3k.

Step 3: Because y = x 2? Tangent to y =KX+b, so there is only one intersection point, that is, the simultaneous equation of these two equations has only one solution (△=0).

Bring y = x 2 into x 2 = kx+b of Y=kx+b, and then x 2 = kx+5-3k? Let △=0 to get K=2 or K= 10. When K=2? When b=- 1K= 10? b=-25

Step 4: So the tangent equation passing through p(3.5) and y = x 2 is: Y=2X- 1 or Y= 10X-25.

2:

As shown in the figure, AB=AF+BE (the distance from the point on the parabola to the focus is equal to the distance from the directrix).

Because the slope of parabolic focus (1.0) is k= 1, the linear equation is: y=x- 1 and y 2 = 4x, and x = 3 2 √ 2 is obtained.

Available graphs AF+BE=X 1+X2+P=8.