parabola

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christy

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Consider a parabola y=x^(2)-4x+8.
(a) Write down two straight lines that touch the parabola at only one point.
(b) Find the corresponding point of contact in (a).
 
christy said:
Consider a parabola y=x^(2)-4x+8.
(a) Write down two straight lines that touch the parabola at only one point.
(b) Find the corresponding point of contact in (a).
Click to expand...
What have you tried? Where are you stuck?

For instance, for Part (a), you started by drawing the parabola. You know from experience that straight lines touching at only one point are "tangent" and must have the same slope as the parabola at that point, so... what did you do with this information?

I'll admit, though, that the parts of this exercise seem a bit backwards. Why would you find the point of tangency after you've picked the point where you're going to find a tangent line? ;)
 
christy said:
Consider a parabola y=x^(2)-4x+8.
(a) Write down two straight lines that touch the parabola at only one point.
(b) Find the corresponding point of contact in (a).
Click to expand...
Are you sure you have copied the problem correctly? There exist an infinite number of lines that "touch the parabola at only one point". If you mean two lines that touch the parabola at only one point and the same point, there are no such lines. A parabola has a unique tangent at every point.
 
HallsofIvy said:
Are you sure you have copied the problem correctly? There exist an infinite number of lines that "touch the parabola at only one point". If you mean two lines that touch the parabola at only one point and the same point, there are no such lines. A parabola has a unique tangent at every point.
Click to expand...

i don't know why the answer is y=4 and y=2x-1.
 
Perhaps you "don't know why the answer is y=4 and y=2x-1" because it is NOT! Did you notice that \(\displaystyle x^2- 4x+ 4= (x- 2)^2\). The line y= 4 crosses that when \(\displaystyle y= 4= (x- 2)^2\) which is the same as \(\displaystyle x- 2= \pm 4\) and so crosses the parabola at (2, 4) and (-2, 4). y= 4 does NOT touch the parabola at only one point. \(\displaystyle y= 2x- 1= x^2- 4x+ 4\) is the same as \(\displaystyle x^2- 6x+ 5= (x- 5)(x- 1)= 0\). That line crosses the parabola at (5, 9) and (1, 1). It does NOT touch the parabola at only one point.
 
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