- #1

- 5

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter swechan02
- Start date

- #1

- 5

- 0

- #2

- 793

- 4

- #3

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

Jameson is correct but it's a bit more complicated than he implies.

Suppose a line is tangent to y= x^{2}+ 1 at (x_{0},x_{0}^{2}+ 1) and tangent to y= -x^{2} at (x_{1},-x_{1}^{2}).

Any (non-vertical) line can be written as y= mx+ b. m is equal to the derivative of the functions at the given points: m= 2x_{0}= -2x_{1} so x_{1}= -x_{0}. We must have x_{0}^{2}+1= (2x_{0})x_{0}+ 1 or b= 1- x_{0}^{2}. We must also have -x_{1}^{2}= (-2x_{1})x_{1}+ b or b= x_{1}^{2}. That is, b= x_{1}^{2}= 1- x_{0}^{2}. But since x_{1}= -x_{0}, x_{1}^{2}= x_{0}^{2} so 1- x_{0}^{2}= x_{0}^{2}.

Solve that for x_{0} and then you can find m and b.

Because of the squares, there are, of course, two symmetric solutions,.

Suppose a line is tangent to y= x

Any (non-vertical) line can be written as y= mx+ b. m is equal to the derivative of the functions at the given points: m= 2x

Solve that for x

Because of the squares, there are, of course, two symmetric solutions,.

Last edited by a moderator:

Share: