October 12th, 2014, 08:25 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+(x1)^2 . Last edited by skipjack; October 13th, 2014 at 09:04 AM. 
October 12th, 2014, 09:17 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,980 Thanks: 1573 
a line tangent to the curve $\displaystyle y = 1+(x1)^2$ will pass thru the points $\displaystyle (0,0)$ and $\displaystyle \left(x,1+(x1)^2\right)$ ... slope of any such line is $\displaystyle \frac{\Delta y}{\Delta x} = \frac{1+(x1)^2  0}{x0} = \frac{1+(x1)^2}{x}$ now ... what do you know about the slope of a tangent line from a calculus perspective? 
October 12th, 2014, 02:31 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond 
$\displaystyle y=kx$ $\displaystyle y=1+(x1)^2$ $\displaystyle k=2x2$ $\displaystyle (2x2)x=1+(x1)^2\Rightarrow x=\pm\sqrt2$ $\displaystyle y=2(\sqrt21)x$ $\displaystyle y=2(\sqrt2+1)x$ 
October 13th, 2014, 07:38 AM  #4  
Newbie Joined: Oct 2014 From: macau Posts: 4 Thanks: 0  Quote:
Quote:
Last edited by skipjack; October 13th, 2014 at 09:22 AM.  
October 13th, 2014, 09:21 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,820 Thanks: 2159 
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. 

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