# Find tangent lines to both curves

## Homework Statement

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

## 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
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## Homework Statement

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

## 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.

Gosh I am so blind, thank you so much!