#### To determine

**To find:** The number of lines that are normal to the parabola y=x2 such that it passes through the point (0,c) when c>12 and c≤12.

#### Answer

Three normal lines to the parabola that passes through the point (0,c) if c>12.

One normal line to the parabola that passes through the point (0,c) if c≤12.

#### Explanation

**Derivative rule:**

Power rule: ddx(xn)=nxn−1

**Calculation:**

The derivative of the curve y=x2 is computed as follows.

dydx=ddx(x2)

Apply the power rule (1) and simplify the expression,

dydx=2x2−1=2x

Thus, the derivative of the curve is dydx=2x.

Therefore, the slope of the tangent line to the curve is 2*x*.

Obtain the slope of the normal line to the curve by using the slope of the tangent line.

Since every point of the parabola is of the form (a,a2), the slope of the tangent to the curve at (a,a2) is 2*a*.

Here, the tangent line is perpendicular to the normal line. That is, if m1 and m2 are the slopes of tangent line and normal line, then m1m2=−1.

This implies that, the slope of the normal line to the curve is −12a.

Note that, the slope of the line passing through the points (x1,x2) and (y1,y2) is m=y2−y1x2−x1.

Here, the normal line passing through the points (0,c) and (a,a2).

The slope of the normal line is computed as follows.

m=a2−ca−0=a2−ca

Since the slope of normal line is −12a, the equation becomes,

−12a=a2−ca−12=a2−ca2=c−12

The above equation has two solution if c>12, one solution if c=12 and no solution if c<12.

Since the *y*-axis is normal to the parabola and passes through the point (0,c) (independent form *c*).

Therefore, there are three normal lines to the parabola that passes through the point (0,c) if c>12 and one normal line to the parabola that passes through the point (0,c) if c≤12.