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October 12th, 2014, 08:25 AM   #1
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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 09:04 AM.
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October 12th, 2014, 09:17 AM   #2
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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?
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October 12th, 2014, 02:31 PM   #3
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$\displaystyle y=kx$

$\displaystyle y=1+(x-1)^2$

$\displaystyle k=2x-2$

$\displaystyle (2x-2)x=1+(x-1)^2\Rightarrow x=\pm\sqrt2$

$\displaystyle y=2(\sqrt2-1)x$

$\displaystyle y=-2(\sqrt2+1)x$
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October 13th, 2014, 07:38 AM   #4
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Quote:
Originally Posted by greg1313 View Post
$\displaystyle y=kx$

$\displaystyle y=1+(x-1)^2$

$\displaystyle k=2x-2$

$\displaystyle (2x-2)x=1+(x-1)^2\Rightarrow x=\pm\sqrt2$

$\displaystyle y=2(\sqrt2-1)x$

$\displaystyle y=-2(\sqrt2+1)x$
thanks sensei !!! that is very helpful~~



Quote:
Originally Posted by skeeter View Post
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 09:22 AM.
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October 13th, 2014, 09:21 AM   #5
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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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