Question

Sketch the parabolas \(y = {x^{\bf{2}}}\) and \(y = {x^{\bf{2}}} - {\bf{2}}x + {\bf{2}}\). Do you think there is a line that is tangent to both curves? If so, find its equation. If not, why not?

Short answer

The equation of common tangent is \(y = x - \frac{1}{4}\).

Step-by-step solution

Make the sktech of the given parabolas

Use the following steps to plot the graph of given functions:

1. In the graphing calculator select “STAT PLOT” and enter the equations \({x^2}\) and \({x^2} - 2x + 2\).
2. Enter the graph button in the graphing calculator.

The figure below repsents the parabolas \(y = {x^2}\) and \(y = {x^2} - 2x + 2\) and a possible common tangents to the curve.

Find the equation of common tangent

Let the point \(P\left( {a,{a^2}} \right)\) on the curve \(y = {x^2}\) and \(Q\left( {b,\,{b^2} - 2b + 2} \right)\) is on the parabola \(y = {x^2} - 2x + 2\).

The slope of the line joining the points \(\left( {a,{a^2}} \right)\) and \(\left( {b,{b^2} - 2b + 2} \right)\) is:

\(m = \frac{{{b^2} - 2b + 2 - {a^2}}}{{b - a}}\)

The slope of tangent to the curve \(y = {x^2}\) is \(2x\). Therefore, for the point \(\left( {a,{a^2}} \right)\) is:

\(m = 2a\)

The slope of tangent to the curve \(y = {x^2} - 2x + 2\) is \(2x - 2\). Therefore, for the point \(\left( {b,{b^2} - 2b + 2} \right)\):

\(m = 2b - 2\)

Find the values of a and b

The value of three slope calculated in step 2 must be equal.

\(\begin{aligned}2a &= 2b - 2\\a &= b - 1\end{aligned}\)

And

\(\begin{aligned}2b - 2 &= \frac{{{b^2} - 2b + 2 - {a^2}}}{{b - a}}\\2b - 2 &= \frac{{{b^2} - 2b + 2 - {{\left( {b - 1} \right)}^2}}}{{b - \left( {b - 1} \right)}}\\2b - 2 &= {b^2} - 2b + 2 - {b^2} + 2b - 1\\2b &= 3\\b &= \frac{3}{2}\end{aligned}\)

So, \(a = \frac{1}{2}\)

So, the point of tangents are:

\(\left( {a,{a^2}} \right) \equiv \left( {\frac{1}{2},\frac{1}{4}} \right)\)

And

\(\begin{aligned}\left( {b,{b^2} - 2b + 2} \right) &\equiv \left( {\frac{3}{2},{{\left( {\frac{3}{2}} \right)}^2} - 2\left( {\frac{3}{2}} \right) + 2} \right)\\ &\equiv \left( {\frac{3}{2},\frac{9}{4} - 1} \right)\\ &\equiv \left( {\frac{3}{2},\frac{5}{4}} \right)\end{aligned}\)

Find the equation of common tangent

The equation of line tangent to \(\left( {\frac{1}{2},\frac{1}{4}} \right)\) and \(\left( {\frac{3}{2},\frac{5}{4}} \right)\) is:

\(\begin{aligned}y - \frac{1}{4} &= 2\left( {\frac{1}{2}} \right)\left( {x - \frac{1}{2}} \right)\\y - \frac{1}{4} &= x - \frac{1}{2}\\y &= x - \frac{1}{4}\end{aligned}\)

So, the equation of common tangent is \(y = x - \frac{1}{4}\).