#### To determine

**To sketch:** The parabolas y=x2 and y=x2−2x+2 and check whether there is a tangent line to both the curves or not.

#### Explanation

**Derivative rules:**

(1) Power Rule: ddx(xn)=nxn−1

(2) Constant multiple rule: ddx(cf)=cddx(f)

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

(4) Difference rule: ddx(f−g)=ddx(f)+−ddx(g)

**Result used:**

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

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

**Graph:**

The graph of two parabolas y=x2 and y=x2−2x+2 is shown below in Figure 1.

From Figure 1, it is observed that there may be a line that is tangent to both the parabolas.

It is required to find the equation of the tangent line to the parabolas.

**Calculation:**

Consider the parabolas y=x2 and y=x2−2x+2.

Choose the point *P*(a,a2) and *Q*(b,b2−2b+2) on the parabolas y=x2 and y=x2−2x+2, respectively.

Suppose the slope of the required tangent line passes through the points *P*(a,a2) and *Q*(b,b2−2b+2), then m=b2−2b+2−a2b−a. (2)

The derivative of parabola y=x2 is dydx, which is obtained as follows.

dydx=ddx(x2)

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

dydx=2x

Thus, the derivative of y=x2 is 2*x*.

Therefore, the slope of the tangent to y=x2 at (a,a2) is 2*a*. (3)

The derivative of parabola y=x2−2x+2 is dydx, which is obtained as follows.

dydx=ddx(x2−2x+2)

Apply the derivative rules (1), (2), (3) and (4),

dydx=ddx(x2)−ddx(2x)+ddx(2)=ddx(x2)−2ddx(x)+ddx(2)=2x−2

Thus, the derivative of y=x2−2x+2 is 2x−2.

Therefore, the slope of the tangent to y=x2−2x+2 at (b,b2−2b+2) is 2b−2. (4)

Since the required equation of the tangent is linear from (a,a2) to (b,b2−2b+2), the slopes of tangent line are equal.

From equations (2) and (3), 2b−2=b2−2b+2−a2b−a.

From equations (3) and (4), a=b−1.

Substitute a=b−1 in 2b−2=b2−2b+2−a2b−a.

2b−2=b2−2b+2−(b−1)2b−(b−1)2b−2=b2−2b+2−(b2+1−2b)2b−2=b2−2b+2−b2−1+2b2b−2=1

Add 2 on both sides and obtain the value of *b.*

2b=1+22b=3b=32

Substitute the value b=32 in a=b−1,

a=32−1a=12

For a=12, the slope 2*a* is 1 and the point *P*(a,a2) is (12,14).

Substitute (12,14) for (x1,y1) and 1 for *m* in equation (1),

y−14=1(x−12)y=x−12+14y=x−24+14y=x−14

Therefore, the equation of the tangent line to the parabolas is y=x−14.