Finding all tangent lines through the origin.

Limits, differentiation, related rates, integration, trig integrals, etc.
sepoto
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Finding all tangent lines through the origin.

$\mbox{Question 1C-5:}$
. . .$\mbox{Find all tangent lines through the origin}$
. . .$\mbox{to the graph of }\, y\, =\, 1\, +\, (x\, -\, 1)^2$

$\mbox{Solution 1C-5:}$
$\mbox{Method 1: }\, y'(x)\, =\, 2(x\, -\, 1),\, \mbox{ so the tangent line through}$
$(a,\, 1\, +\, (a\, -\, 1)^2)\, \mbox{ is }$

. . . . .$y\, =\, 2(a\, -\, 1)(x\, -\, a)\, +\, 1\, +\, (a\, -\, 1)^2$

$\mbox{In order to see if the origin is on this line, plug in }\, x\, =\, 0$
$\mbox{and }\, y\, =\, 0,\, \mbox{ to get the following equation for }\, a:$

. . . . .$0\, =\, 2(a\, -\, 1)(-a)\, +\, 1\, +\, (a\, -\, 1)^2$
. . . . .$=\, -2a^2\, +\, 2a\, +\, 1\, +\, a^2\, -\, 2a\, +\, 1$
. . . . .$=\, -a^2\, +\, 2$

$\mbox{Therefore }\, a\, =\, \pm \sqrt{2}\, \mbox{ and the two tangent lines}$
$\mbox{through the origin are}$

. . . . .$y\, =\, 2(\sqrt{2}\, -\, 1)x\, \mbox{ and }\, y\, =\, -2(\sqrt{2}\, -\, 1)x$

$\mbox{(Because these are lines through the origin, the constant }$
$\mbox{terms must cancel; this is a good check of your algebra!)}$

$\mbox{Method 2: Seek tangent lines of the form }\, y\, =\, mx.\, \mbox{ Suppose}$
$\mbox{that }\, y\, =\, mx\, \mbox{ meets }\, y\, =\, 1\, +\, (x\, -\, 1)^2\, \mbox{ at }\, x\, =\, a;\, \mbox{ then }\, ma\, =$
$1\, +\, (a\, -\, 1)^2.\, \mbox{ In addition, we want the slope }\, y'(a)\, =\, 2(a\, -\, 1)$
$\mbox{to be equal to }\, m,\, \mbox{ so }\, m\, =\, 2(a\, -\, 1).\, \mbox{ Substituting for }\, m,\, \mbox{ we}$
$\mbox{find:}$

. . . . .$2(a\, -\, 1)a\, =\, 1\, +\, (a\, -\, 1)^2$

$\mbox{This is the same equation as Method 1: }\, a^2\, -\, 2\, =\, 0,\, \mbox{ so}$
$a\, \pm\, \sqrt{2}\, \mbox{ and }\, m\, =\, 2(\pm \sqrt{2}\, -\, 1),\, \mbox{ and the two tangent lines}$
$\mbox{through the origin are as above:}$

. . . . .$y\, =\, 2(\sqrt{2}\, -\, 1)x\, \mbox{ and }\, y\, =\, -2(\sqrt{2}\, -\, 1)x$

To start with I graphed the equation 1+(x-1)^2. I see that the origin of the parabola is shifted one unit up and one unit to the right and the parabola faces upward. So method one applies the formula on a line in point slope form. So an arbitrary point a is applied and x and y are set to zero. The resulting equation is solved for a. I think I understand this far.

What I am still figuring is y=2( (sqrt 2) - 1 )x and y=-2( (sqrt2) + 1 )x

I don't see yet how those were derived fully.

buddy
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Joined: Sun Feb 22, 2009 10:05 pm
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Re: Finding all tangent lines through the origin.

sepoto wrote:$\mbox{Method 1: the tangent line is }$
. . . . .$y\, =\, 2(a\, -\, 1)(x\, -\, a)\, +\, 1\, +\, (a\, -\, 1)^2$
$a\, =\, \pm \sqrt{2}$

plug sqrt[2] in for a in 1st line eqn:
y=2(sqrt[2]-1)(x-sqrt[2])+1+(sqrt[2]-1)^2
=2(sqrt[2]x-x-2+sqrt[2])+1+(2-2sqrt[2]+1)
=2sqrt[2]x-2x-4+2sqrt[2]+1+2-2sqrt[2]+1
=2sqrt[2]x-2x-4+1+2+1+2sqrt[2]-2sqrt[2]
=2sqrt[2]x-2x
=2(sqrt[2]x-x)
=2(sqrt[2]-1)x
the other one works the same way