(a)

The equation of parabola is   and the point is .

Slope of the tangent is derivative of the curve.

.

Apply derivative on each side with respect to .

Slope of the tangent  is .

Point-slope form of line equation : .

Substitute  and  in the above formula.

This is a pair of tangent lines.

These tangent lines intersect the parabola, and the intersecting points can be determined  by solving them.

Substitute  in the curve .

 and .

Substitute  values in .

If , then .

If , then .

Therefore, the points at tangent lines intersect parabola are  and .

Tangent line passing through :

.

Find the slope at .

.

Point-slope form of line equation : .

Substitute  and  in the above formula.

Tangent line  passing through :

.

Find the slope at .

.

Point-slope form of line equation : .

Substitute  and  in the above formula.

(b)

The equation of parabola is  and the point is .

Slope of the tangent to parabola is .

At  , .

Assume that at  is the tangent point.

Slope of the tangent line at  is .

Slope of the line passing through two points  and  is defined as .

Here  and 

Discriminant of quadratic equation  is .

Here  and .

Since the discriminant is negative there is no real values of .

Therefore there is no tangent line at .

Graph:

Graph the curve with the point .

Tangents are  and .

Graph is

.