Finding all tangent lines through the origin.  TOPIC_SOLVED

Limits, differentiation, related rates, integration, trig integrals, etc.

Finding all tangent lines through the origin.

Postby sepoto on Tue Oct 29, 2013 5:36 am


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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.

Thanks in advance...
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Re: Finding all tangent lines through the origin.  TOPIC_SOLVED

Postby buddy on Tue Oct 29, 2013 10:52 am

sepoto wrote:
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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
:wave:
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