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 October 22nd, 2013, 09:59 AM #1 Member   Joined: Oct 2013 Posts: 31 Thanks: 0 Common tangents to circle and parabola Hello! Problem: Find the common tangents to the circle x^2+y^2=1 and the parabola y=x^2+1. The tangents are y=1 (found that), y=2*sqrt(6)x-5 and y=-2*sqrt(6)x-5. It's "supposed" to be solved without calculus but any reliable method is appreciated! Thanks
 October 22nd, 2013, 10:40 AM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Common tangents to circle and parabola The circle and parabola share only one common point, which is the point $(x, y)= (0, 1)$. By an easy graphical sketch, you can tell that they share a common tangent at this point, which is the line $y= 1$.
 October 22nd, 2013, 10:51 AM #3 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Common tangents to circle and parabola The method I used to get the common tangents was: Choose an arbitrary point (a, b) on the parabola. The tangent at this point is $y= 2ax - 2a^2 + b = 2ax -a^2 + 1$ Then look for (x,y) that lies on this tangent with $x^2 + y^2= 1$ This gives a quadratic in x: $(4a^2+1)x^2 + 4a(1-a^2)x + (1-a^2)^2 - 1= 0$ Now, we are looking for the situation where this quadratic has precisely one real solution (as this equates to the tangent touching the circle, rather than missing it or cutting it with two points of intersection). So, we need the determinant of this quadratic to be zero. Hence: $(16a^2)(1-a^2)^2= 4(4a^2+1)[(1-a^2)^2 - 1]$ This gives: $a= 0 \ or \ a = \pm \sqrt{6}$ You can then plug a back into the equation for the tangent to the parabola.
 October 22nd, 2013, 10:58 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,948 Thanks: 1139 Math Focus: Elementary mathematics and beyond Re: Common tangents to circle and parabola For the lines to be tangent, the discriminants of the equations below (on the left) must equal zero (m is slope and b is y-intercept). $x^2\,+\,1\,=\,mx\,+\,b\,\Rightarrow\,m^2\,+\,4(b\,-\,1)\,=\,0\;[1]$ $-\sqrt{1\,-\,x^2}\,=\,mx\,+\,b\,\Rightarrow\,b\,=\,\sqrt{m^2\ ,+\,1}\;[2]$ Now sub the result of [2] into [1]: $m^2\,+\,4$$\sqrt{m^2\,+\,1}\,-\,1$$\,=\,0\,\Rightarrow\,m\,=\,\pm2\sqrt6$ and I'll let you finish up from there.
 October 22nd, 2013, 12:11 PM #5 Member   Joined: Oct 2013 Posts: 31 Thanks: 0 Re: Common tangents to circle and parabola Thanks alot Greg. I had similar reasoning, and put (a,b) as a point on the circle and then got the slope of the tangent line to the circle to -b/a. Couldn't solve anything though since I didn't realize b=sqrt(1-a^2), haha. And well done Pero, altough that gets a bit hairy for my taste! :P
 October 22nd, 2013, 12:58 PM #6 Senior Member   Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116 Re: Common tangents to circle and parabola It might be worth making sure you understand Greg's solution as it may be cleverer than you think. And even when dealing with circles and parabolas, he likes to be elliptical! Where did those equations come from?
 October 22nd, 2013, 08:05 PM #7 Member   Joined: Oct 2013 Posts: 31 Thanks: 0 Re: Common tangents to circle and parabola As I understand it, it's saying that we have a line that intersects a circle and a parabola, and by setting the discriminant to zero we make sure that it intersects both in a single point respectively. (discriminant has to be zero if the quadratic has one solution) The second equation should also generate m=0 which gives b=1, and you get the y=1 part.
 October 22nd, 2013, 08:27 PM #8 Member   Joined: Oct 2013 Posts: 31 Thanks: 0 Re: Common tangents to circle and parabola Well, whatever equation you substitute into.
October 23rd, 2013, 02:37 AM   #9
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Quote:
 Originally Posted by greg1313 . . . and I'll let you finish up from there.

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