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Two values for m for which a line is tangent to a parabola

  • Thread starter rakeru
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Homework Statement



Hi! I am doing some problems to practice for a math competition, and I'm wondering if I did this correctly. I don't really have an answer sheet, so I have no way of knowing whether I'm right. If you would please review it, that would be cool!

It reads:

There are exactly two values of m for which the line y=mx-12 is tangent to the parabola y=x^2-3x+5. Determine those two values.

Homework Equations



y=mx-12
y=x^2-3x+5

The Attempt at a Solution



Okay so first I found the derivative of y=x^2-3x+5.

y'=2x-3

This is the slope, so I put it into the equation of the tangent line:

y=mx-12=(2x-3)x-12=2x^2-3x-12 <-- the tangent line

Now I made the the two equations equal to each other because they intersect at one point, right?

so:

x^2-3x+5=2x^2-3x-12

and when I solve for x it gives me +/- √17

I put this into the equation for the slope and I get:

m= 2√17 -3 or m=-2√17 -3

Is this correct? Thank you in advance.
 

Answers and Replies

  • #2
Nathanael
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Looks correct to me
 
  • #3
HallsofIvy
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Another way to do it (used by Fermat previous to the development of calculus):
To find where the line y= mx- 12 crosses the parabola [itex]y= x^2- 2x+ 5[/itex], solve the equation [itex]mx- 12= x^2- 2x+ 5[/itex] or [itex]x^2- (m+ 3)x+ 17= 0[/itex]

The "discriminant" of that equation is [itex](m+3)^2- 4(1)(17)= m^2+ 6m+ 9- 68= m^2+ 6x- 59[/itex]. The line will not touch the parabola at all if the equation has no real roots- if the discriminant is negative. It will cross through the parabola if the equation has two real roots- if the discriminant is positive. It will touch the parabola once (be tangent to the parabola) if the equation has a single root- if the discriminant is zero.

Thus, the line y= mx- 12 will be tangent to the parabola [itex]y= x^2- 2x+ 5[/itex] if and only if [itex]m^2+ 6m- 59= 0[/itex]. That is the same as the equation you got and, yes, it has roots [itex]-3\pm 2\sqrt{17}[/itex].
 
  • #4
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  • #5
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Another way to do it (used by Fermat previous to the development of calculus):
To find where the line y= mx- 12 crosses the parabola [itex]y= x^2- 2x+ 5[/itex], solve the equation [itex]mx- 12= x^2- 2x+ 5[/itex] or [itex]x^2- (m+ 3)x+ 17= 0[/itex]

The "discriminant" of that equation is [itex](m+3)^2- 4(1)(17)= m^2+ 6m+ 9- 68= m^2+ 6x- 59[/itex]. The line will not touch the parabola at all if the equation has no real roots- if the discriminant is negative. It will cross through the parabola if the equation has two real roots- if the discriminant is positive. It will touch the parabola once (be tangent to the parabola) if the equation has a single root- if the discriminant is zero.

Thus, the line y= mx- 12 will be tangent to the parabola [itex]y= x^2- 2x+ 5[/itex] if and only if [itex]m^2+ 6m- 59= 0[/itex]. That is the same as the equation you got and, yes, it has roots [itex]-3\pm 2\sqrt{17}[/itex].

Ouuuu interesting. Thank you :)
 

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