Find tangent lines to both curves

[drawar]

Homework Statement

Find the equation of all straight lines, if any, that are tangent to both the curves y = {x^2} + 4x + 1 and y = - {x^2} + 4x - 1.

Homework Equations

The Attempt at a Solution

Suppose such a line exists and its slope is m. Let ({x_1},{y_1}) and ({x_2},{y_2}) be the tangent points on the curves y = {x^2} + 4x + 1 and y = - {x^2} + 4x - 1 respectively.
Then {y_1} = {x_1}^2 + 4{x_1} + 1 and {y_2} = - {x_2}^2 + 4{x_2} - 1.
The slope of the curves at {x_1} is 2{x_1} + 4 and at {x_2} is -2{x_2} + 4. Thus m=2{x_1} + 4=-2{x_2} + 4 \Rightarrow {x_1} = - {x_2}. Moreover, m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{( - {x_2}^2 + 4{x_2} - 1) - ({x_1}^2 + 4{x_1} + 1)}}{{{x_2} - {x_1}}} = \frac{{{y_1}}}{{{x_1}}}.
What should be the next steps here? Thanks!

 

[SammyS]

Homework Statement

Find the equation of all straight lines, if any, that are tangent to both the curves y = {x^2} + 4x + 1 and y = - {x^2} + 4x - 1.

Homework Equations

The Attempt at a Solution

Suppose such a line exists and its slope is m. Let ({x_1},{y_1}) and ({x_2},{y_2}) be the tangent points on the curves y = {x^2} + 4x + 1 and y = - {x^2} + 4x - 1 respectively.
Then {y_1} = {x_1}^2 + 4{x_1} + 1 and {y_2} = - {x_2}^2 + 4{x_2} - 1.
The slope of the curves at {x_1} is 2{x_1} + 4 and at {x_2} is -2{x_2} + 4. Thus m=2{x_1} + 4=-2{x_2} + 4 \Rightarrow {x_1} = - {x_2}. Moreover, m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{( - {x_2}^2 + 4{x_2} - 1) - ({x_1}^2 + 4{x_1} + 1)}}{{{x_2} - {x_1}}} = \frac{{{y_1}}}{{{x_1}}}.
What should be the next steps here? Thanks!

You have it nearly done.

Well, you have:

x_1=-x_2

m=-2{x_2} + 4

\displaystyle m=\frac{{( - {x_2}^2 + 4{x_2} - 1) - ({x_1}^2 + 4{x_1} + 1)}}{{{x_2} - {x_1}}}​

Plug -x₂ in for x₁ in that last equation & equate that to your other expression for m that has only x₂ in it.

Solve for x₂.

 

[drawar]

Gosh I am so blind, thank you so much!