#### To determine

**To find:** The second point of intersection.

#### Answer

The normal line intersect the parabola second time at (32,54).

#### Explanation

**Given:**

The normal line intersect at the parabola y=x2−1 at two points.

The first point of intersection at (−1,0).

**Derivative rules:**

(1) Constant Function: ddx(c)=0

(2) Power Rule: ddx(xn)=nxn−1

(3) Difference Rule: ddx[f(x)−g(x)]=ddx(f(x))−ddx(g(x))

**Formula used:**

The equation of normal line at (x1,y1) is, y−y1=−1m(x−x1) (1)

Here, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

**Calculation:**

The derivative of y is dydx, which is obtained as follows,

dydx=ddx(y) =ddx(x2−1)

Apply the derivative rule (3) and (1),

dydx=ddx(x2)−ddx(1) =ddx(x2)−0

Apply the power rule (2) and simplify the expression.

dydx=2x2−1=2x

Therefore, the derivative of the function is dydx=2x.

Obtain the equation of the normal line to curve y=x2−1 at the point (−1,0).

The slope of the tangent line at (−1,0) is,

m=dydx|x=−1=2(−1)=−2

Thus, the slope of the tangent line at (−1,0) is m=−2.

Substitute (−1,0) for (x1,y1) and −2 for *m* in equation (1),

y−0=−1−2(x−(−1))y−0=12(x+1)y=x2+12y=12(x+1)

Therefore, the equation of the normal line to the parabola y=x2−1 at (−1,0) is y=12(x+1).

Obtain second point of intersection.

Consider the equations y=x2−1 and y=12(x+1).

Here, the left-hand sides of the equations are same, the right-hand side of the equations must be same.

x2−1=12(x+1)

Cross multiply and simplify the expression,

2(x2−1)=x+12x2−2=x+12x2−2−x−1=02x2−x−3=0

Use the quadratic formula and obtain the values of *x*.

x=−(−1)±(−1)2−4(2)(−3)2(2)=1±1+244=1±254=1±54

Thus, the values of *x* are x=−1 and x=32.

Substitute x=32 in normal equation y=12(x+1),

y=12(32+1)=12(3+22)=12(52)=54

Therefore, the normal line intersect the parabola second time at (32,54).

**Graph:**

The graph of the parabola and the normal line is shown below in Figure 1.

From Figure 1, it is observed that the line y=12(x+1) is normal to the parabola at the point (−1,0) and (32,54).