The question asks to find the equation of the line passing through $(0,2)$ and just touching $y=(x+2)^{2}$ where $x>0$

I let the equation of the line have gradient $m$ therefore the line is $y=mx+2$ and for the lines to meet: $$mx+2=(x+2)^{2}$$ $$\therefore x^{2}-(4+m)x+2=0$$

I then let the discriminant equal zero and solved for $m$ to get: $$m=-4\pm\sqrt{2}$$ But $x>0$ therefore $m=-4+\sqrt{2}$

I then questioned this as to why I let the discriminant equal zero, and what does this mean. I then did it another way by again letting $mx+2=x^{2}-4x+4$ and solving this for $x$ by substitution with the equation $m=2x-4$ (their derivatives with respect to $x$ as the lines should only touch I realised the gradient at the points where they did would be equal). I again took the positive root as $x>0$ and again got $m=-4+\sqrt{2}$. My question is first of all is this correct and if so why did letting the discriminant be equal to zero give me those specific roots that I was looking for? Thanks

  • $\begingroup$ A discriminant of zero corresponds to our line touching the parabola at exactly one point. Consider that the quadratic formula then gives only one zero. $\endgroup$ – obataku May 29 '13 at 0:01
  • $\begingroup$ Oh so I got two solutions where two lines touch $y=(x+2)^{2}$ at one point? $\endgroup$ – Johnmgee May 29 '13 at 0:02
  • 2
    $\begingroup$ those are the two possible slopes of lines that cross $(0,2)$ while only touching our parabola at a point. One of them is where $x<0$ while another is where $x>0$. :-) $\endgroup$ – obataku May 29 '13 at 0:05
  • $\begingroup$ Yeah I think I get it, thanks $\endgroup$ – Johnmgee May 29 '13 at 0:05

As @oldrinb pointed out in the comments, setting the discriminant equal to zero corresponds to having the line touch the parabola at only one point. This is because having them touch only once means that there is only one valid value for $x$ to the equation $x^{2}-(4+m)x+2=0$.

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