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Topic: Tangent through the origin. (Read 774 times) |
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Benoit_Mandelbrot
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Tangent through the origin.
« on: Jan 16th, 2004, 11:51am » |
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A straight line through the origin, y=mx, is tangent to the parabola y=x2-7x+17 at (x0,y0). Find x0 and m, and explain how you got it.
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John_Gaughan
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Re: Tangent through the origin.
« Reply #1 on: Jan 16th, 2004, 12:02pm » |
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I don't have time to solve it right at this moment but I have a general idea. :: The parabola opens up, since a is positive. By taking the derivative, its minimum point is at (3.5, 4.75). x0 is roughly 3.5, and m is roughly 1.36. I could solve it exactly but I gotta go now... I'll work on it later. ::
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ThudnBlunder
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Re: Tangent through the origin.
« Reply #2 on: Jan 16th, 2004, 2:28pm » |
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When has this got to be in by?
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Sameer
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Re: Tangent through the origin.
« Reply #3 on: Jan 16th, 2004, 2:53pm » |
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Now is there a catch to this... i used to solve this in high school...
differentiate equation to get the slope of the tangent
y' = 2*x - 7
Since this is tangent at (xo, yo) point we have
y' = 2*xo - 7 = m ..... (i)
Also the point lies on the curve so
yo = xo^2 - 7xo + 17 .... (ii)
Also this point is on the tangent line so
yo = m*xo .... (iii)
using these equations
m*xo = xo^2 - 7xo + 17
(2*xo-7)*xo = xo^2 - 7xo + 17
so xo^2 = 17
or xo = +/- sqrt(17)
thus m = +/- 2*sqrt(17) - 7
Thus we have two tangents and two points on the parabola that satisfy the condition (as expected)
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ThudnBlunder
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Re: Tangent through the origin.
« Reply #4 on: Jan 16th, 2004, 3:38pm » |
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We have y = mx = x2 - 7x + 17 when line and curve meet.
So x2 - (m + 7)x + 17 = 0
[smiley=therefore.gif] x = [m + 7 [smiley=pm.gif][smiley=surd.gif]{(m + 7)2 - 68}]/2
As the line is a tangent the discriminant equals zero.
Hence m2 + 14m - 19 = 0
m = -7 [smiley=pm.gif] 2[smiley=surd.gif]17
and 2x = m + 7
So x0 = [smiley=pm.gif][smiley=surd.gif]17
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| « Last Edit: Jan 16th, 2004, 3:42pm by ThudnBlunder » |
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Sameer
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Re: Tangent through the origin.
« Reply #5 on: Jan 16th, 2004, 4:01pm » |
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Is this one of the "do not post by math phds" type of post? 
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"Obvious" is the most dangerous word in mathematics.
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Proof is an idol before which the mathematician tortures himself.
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Icarus
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Re: Tangent through the origin.
« Reply #6 on: Jan 18th, 2004, 11:44am » |
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on Jan 16th, 2004, 4:01pm, Sameer wrote:| Is this one of the "do not post by math phds" type of post? |
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Now I feel discriminated against! 
Actually, the approach T&B takes is the same as yours, but tightened up a bit. Personally, my idea on how to solve it was exactly the same as yours, but since I knew it would work, I just kept reading to see who actually came up with it!
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Benoit_Mandelbrot
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Re: Tangent through the origin.
« Reply #7 on: Jan 19th, 2004, 6:05am » |
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Good work. What I did is I found the equation of the tangent line, and since the y-intercept must be zero, I set the y-intercept equal to zero and solved for x0. I got a quadratic x02=7, and I got the same answers as you all did.
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Sameer
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Re: Tangent through the origin.
« Reply #8 on: Jan 19th, 2004, 6:39am » |
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on Jan 18th, 2004, 11:44am, Icarus wrote:
Now I feel discriminated against!
Actually, the approach T&B takes is the same as yours, but tightened up a bit. Personally, my idea on how to solve it was exactly the same as yours, but since I knew it would work, I just kept reading to see who actually came up with it! |
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LOL Icarus.. I will let the stars against your name speak out... 
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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Benoit_Mandelbrot
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Re: Tangent through the origin.
« Reply #9 on: Jan 19th, 2004, 11:53am » |
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on Jan 16th, 2004, 2:28pm, THUDandBLUNDER wrote:| When has this got to be in by? |
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What are you talking about? There is no deadline for this.
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towr
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Re: Tangent through the origin.
« Reply #10 on: Jan 19th, 2004, 12:38pm » |
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I'm sure he was just kidding, but some people do try to make us do their homework :p
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SWF
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Re: Tangent through the origin.
« Reply #11 on: Jan 19th, 2004, 5:14pm » |
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Here is an alternative approach for those who understand parabolas, but maybe not derivatives:
y=x2-2ax+17=(x-a)2+(17-a2)
From this "standard form" of the equation it is seen that the focus, F, is at (a,v+0.25) and directrix has y coordinate v-0.25, where v is the y-coordinate of the parabola's vertex. The origin O is at (0,0), and a point of tangency P is at (x,x2-2ax+17). Let D be the projection of P on the directrix, at (x,v-0.25).
By definition of parabola, triangle FPD is isosceles. Also OP is perpendicular to DF (evident from either the reflection property of parabolas, or can be seen by considering how to find points on parabola adjacent to P). Dot product of DF with OP is therefore zero:
0=(a-x)*x + 0.5*(x2-2ax+17)
x2=17 (independent of values for a and v, although slope depends on a).
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Benoit_Mandelbrot
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Re: Tangent through the origin.
« Reply #12 on: Jan 20th, 2004, 8:41am » |
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That's the thing. Usually there are many more than one way to solve a problem. Some are easier than others. It all depends on how we see the problem and how we see the solution that's easier for us.
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