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        riddles >> easy >> Tangent through the origin.
        
(Message started by: Benoit_Mandelbrot on Jan 16^th, 2004, 11:51am)

Title: Tangent through the origin.
Post by Benoit_Mandelbrot on Jan 16^th, 2004, 11:51am

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.

Title: Re: Tangent through the origin.
Post by John_Gaughan on Jan 16^th, 2004, 12:02pm

I don't have time to solve it right at this moment but I have a general idea. :: [hide]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.[/hide] ::

Title: Re: Tangent through the origin.
Post by THUDandBLUNDER on Jan 16^th, 2004, 2:28pm

When has this got to be in by?

Title: Re: Tangent through the origin.
Post by Sameer on Jan 16^th, 2004, 2:53pm

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)

Title: Re: Tangent through the origin.
Post by THUDandBLUNDER on Jan 16^th, 2004, 3:38pm

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

Title: Re: Tangent through the origin.
Post by Sameer on Jan 16^th, 2004, 4:01pm

Is this one of the "do not post by math phds" type of post? 8)

Title: Re: Tangent through the origin.
Post by Icarus on Jan 18^th, 2004, 11:44am

on 01/16/04 at 16:01:55, Sameer wrote:

Is this one of the "do not post by math phds" type of post? 8)

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!

Title: Re: Tangent through the origin.
Post by Benoit_Mandelbrot on Jan 19^th, 2004, 6:05am

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.

Title: Re: Tangent through the origin.
Post by Sameer on Jan 19^th, 2004, 6:39am

on 01/18/04 at 11:44:51, Icarus wrote:

LOL Icarus.. I will let the stars against your name speak out... ;D

Title: Re: Tangent through the origin.
Post by Benoit_Mandelbrot on Jan 19^th, 2004, 11:53am

on 01/16/04 at 14:28:55, THUDandBLUNDER wrote:

When has this got to be in by?

What are you talking about? There is no deadline for this.

Title: Re: Tangent through the origin.
Post by towr on Jan 19^th, 2004, 12:38pm

I'm sure he was just kidding, but some people do try to make us do their homework :p

Title: Re: Tangent through the origin.
Post by SWF on Jan 19^th, 2004, 5:14pm

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).

Title: Re: Tangent through the origin.
Post by Benoit_Mandelbrot on Jan 20^th, 2004, 8:41am

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.