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 October 12th, 2014, 09:25 AM #1 Newbie   Joined: Oct 2014 From: macau Posts: 4 Thanks: 0 About diff. problem I need your solution. ~please!! The question is: find all tangent lines through the origin to the graph of y=1+(x-1)^2 . Last edited by skipjack; October 13th, 2014 at 10:04 AM.
 October 12th, 2014, 10:17 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 a line tangent to the curve $\displaystyle y = 1+(x-1)^2$ will pass thru the points $\displaystyle (0,0)$ and $\displaystyle \left(x,1+(x-1)^2\right)$ ... slope of any such line is $\displaystyle \frac{\Delta y}{\Delta x} = \frac{1+(x-1)^2 - 0}{x-0} = \frac{1+(x-1)^2}{x}$ now ... what do you know about the slope of a tangent line from a calculus perspective? Thanks from topsquark and superzhong
 October 12th, 2014, 03:31 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1156 Math Focus: Elementary mathematics and beyond $\displaystyle y=kx$ $\displaystyle y=1+(x-1)^2$ $\displaystyle k=2x-2$ $\displaystyle (2x-2)x=1+(x-1)^2\Rightarrow x=\pm\sqrt2$ $\displaystyle y=2(\sqrt2-1)x$ $\displaystyle y=-2(\sqrt2+1)x$ Thanks from topsquark and superzhong
October 13th, 2014, 08:38 AM   #4
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Quote:
 Originally Posted by greg1313 $\displaystyle y=kx$ $\displaystyle y=1+(x-1)^2$ $\displaystyle k=2x-2$ $\displaystyle (2x-2)x=1+(x-1)^2\Rightarrow x=\pm\sqrt2$ $\displaystyle y=2(\sqrt2-1)x$ $\displaystyle y=-2(\sqrt2+1)x$
thanks sensei !!! that is very helpful~~

Quote:
 Originally Posted by skeeter a line tangent to the curve $\displaystyle y = 1+(x-1)^2$ will pass thru the points $\displaystyle (0,0)$ and $\displaystyle \left(x,1+(x-1)^2\right)$ ... slope of any such line is $\displaystyle \frac{\Delta y}{\Delta x} = \frac{1+(x-1)^2 - 0}{x-0} = \frac{1+(x-1)^2}{x}$ now ... what do you know about the slope of a tangent line from a calculus perspective?
Thanks, I almost get it. ~XD

Last edited by skipjack; October 13th, 2014 at 10:22 AM.

 October 13th, 2014, 10:21 AM #5 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 Each tangent line through the origin has equation y = kx. If y = kx = 1 + (x - 1)² = x² - 2x + 2, x² - (k + 2)x + 2 = 0. For the required tangents, that quadratic's discriminant must be 0, so k = -2 + 2√2 or -2 - 2√2. Thanks from topsquark

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