Value of when a line touches a parabola at one point.

[mattstjean]

The Question asks:
Find the value of m > 0 for which the line y = mx touches the parabola y = (x - 1)² + 1 at just one point.

So far what I've done...

I know that if the line touches the parabola at one point, it is tangential. So I put together the 2 equations in order to find the intersection (or touch):

mx=(x-1)²+1
0=x²-2x-mx+2
0=x²-(2+m)x+2

That's as far as I've gotten. Someone said I need to find the restriction on 'm' so that the equation only has one root (just touches, as opposed to a full intersection through 2 points), but I don't know what to do...

Any help would be appreciated.

 

[danago]

The Question asks:
Find the value of m > 0 for which the line y = mx touches the parabola y = (x - 1)² + 1 at just one point.

So far what I've done...

I know that if the line touches the parabola at one point, it is tangential. So I put together the 2 equations in order to find the intersection (or touch):

mx=(x-1)²+1
0=x²-2x-mx+2
0=x²-(2+m)x+2

That's as far as I've gotten. Someone said I need to find the restriction on 'm' so that the equation only has one root (just touches, as opposed to a full intersection through 2 points), but I don't know what to do...

Any help would be appreciated.

HINT: Given a quadratic equation ax^2+bx+c=0, its solutions are given by

x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

What should the value of m be so that there is only one solution?

 

[mattstjean]

HINT: Given a quadratic equation ax^2+bx+c=0, its solutions are given by

x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

What should the value of m be so that there is only one solution?

The quadratic formula was my first guess however, since it has two variables: m and x...I didn't know how to use it properly.

 

[Dick]

The quadratic formula was my first guess however, since it has two variables: m and x...I didn't know how to use it properly.

You have x^2-(2+m)x+2=0. A general quadratic equation has the form a*x^2+b*x+c=0. a, b and c depend only on m. They don't depend on x. Just identify a, b and c.

 

[mattstjean]

HINT: Given a quadratic equation ax^2+bx+c=0, its solutions are given by

x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}

What should the value of m be so that there is only one solution?

Is it when b²-4ac = 0 ?

 

[Dick]

Is it when b²-4ac = 0 ?

Yes.