My Math Forum Very Hard Tangent Line Question

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 January 27th, 2009, 10:49 AM #1 Newbie   Joined: Dec 2008 Posts: 2 Thanks: 0 Very Hard Tangent Line Question The green line represents f(x) = (x^2)*sqrt(3) - 1 and red one represents a line which you only know the slope of which is sqrt(3). And for the purpose of my question you can ignore the values in the plot. And my question is how do you compute A (the point at which the red line intercepts the green one) knowing that the red line is tangent to the green line at A, without using a derivative? And without using math more advanced than 11th grade geometry. And here is the plot: http://i134.photobucket.com/albums/q...nsilv/plot.jpg Last edited by skipjack; February 28th, 2018 at 01:16 PM.
 January 27th, 2009, 03:35 PM #2 Senior Member   Joined: Sep 2008 Posts: 116 Thanks: 0 Re: Very Hard Tangent Line Question Sorry, I know this won't be of much help, but discussion sometimes can help... The thing that makes this tricky is that I see three unknowns and only two equations: y = (x^2)*sqrt(3) - 1 y = x * sqrt(3) + b Where x would be your value for a. If it's any help, I can post how to solve it using a derivative, which is the only way I can think of solving it at the moment. f(x) = (x^2)*sqrt(3) - 1 f'(x) = 2x*sqrt(3) sqrt(3) = 2x*sqrt(3) x = 1/2 At the very least, you can use it to check your answer... Last edited by skipjack; February 28th, 2018 at 01:03 PM.
 January 27th, 2009, 07:35 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,934 Thanks: 1128 Math Focus: Elementary mathematics and beyond Re: Very Hard Tangent Line Question $\displaystyle \sqrt{3}x+b=\sqrt{3}x^2-1$ $\displaystyle b=\sqrt{3}x^2-\sqrt{3}x-1$ $\displaystyle b=\sqrt{3}(x-\frac{1}{2})^2-\frac{\sqrt{3}+4}{4}$ ...haven't got any further. Last edited by skipjack; February 28th, 2018 at 12:59 PM.
January 28th, 2009, 03:30 AM   #4
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Quote:
 Originally Posted by shynthriir ... two equations: $\displaystyle y\,=\,(x^2)\sqrt{3}\,-\,1$ $\displaystyle y\,=\,x\sqrt{3}\,+\,b$
$\displaystyle \therefore\,x^2\,-\,x\,=\,\frac{b\,+\,1}{\sqrt{3}}$

$\displaystyle \therefore\,x\,=\,\frac12\,\pm\,\sqrt{\frac{b\,+\, 1}{\sqrt{3}}\,+\,\frac14}\,.$

Since the red line is a tangent, there is only one value for x, and so the square root in the above equation is zero. Hence x = 1/2. You can now find y and b by substituting x = 1/2 into the original equations for y.

Last edited by skipjack; February 28th, 2018 at 01:00 PM.

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