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December 4th, 2009, 08:42 AM   #1
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tangent

Prove that for all values of m, the line $\displaystyle y=mx+\frac{a}{m}$ is a tangent to the parabola $\displaystyle y^2=4ax.$

My attempt:

$\displaystyle y=2a^{\frac{1}{2}}x^{\frac{1}{2}}$

$\displaystyle \frac{dy}{dx}=\sqrt{\frac{a}{x}}$

$\displaystyle m=\sqrt{\frac{a}{x}}$

$\displaystyle a=m^2x$

$\displaystyle y=mx+\frac{m^2x}{m}$

$\displaystyle y=2mx$

I don't know how this is gonna help in my proof ..

Last edited by skipjack; February 28th, 2018 at 03:01 PM.
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December 4th, 2009, 10:36 AM   #2
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Re: tangent

Substitute $\displaystyle y=mx+\frac{a}{m}$ into $\displaystyle y^2=4ax$. You will get a quadratic in x. How many solutions does it have? Solve for x and show that the parabola has slope m at the point of intersection.

Last edited by skipjack; February 28th, 2018 at 02:49 PM.
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December 4th, 2009, 12:59 PM   #3
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Re: tangent

Hello, mikeportnoy!

Quote:
Originally Posted by mikeportnoy View Post
$\displaystyle \text{Prove that for all values of }m\text{,}$

. . $\displaystyle \text{the line }\,y\:=\:mx\,+\,\frac{a}{m}\,\text{ is a tangent to the parabola }\,y^2\:=\:4ax$

aswoods has the best idea . . .

Find the intersection(s) of the line and the parabola.


$\displaystyle \text{Substitute }y \:=\:mx + \frac{a}{m}\text{ into: }\:y^2 \:=\:4ax$

$\displaystyle \text{We have: }\;\left(mx + \frac{a}{m}\right)^2 \:=\:4ax \qquad\Rightarrow\qquad m^2x^2 \,+\, 2ax \,+\, \frac{a^2}{m^2} \:=\:4ax \qquad\Rightarrow\qquad m^2x^2 \,-\, 2ax \,+\, \frac{a^2}{m^2} \:=\:0$

$\displaystyle \text{Multiply by }m^2:\;\;m^4x^2 \,-\, 2am^2x \,+\, a^2 \:=\:0$

$\displaystyle \text{Factor: }\;(m^2x \,-\, a)^2 \:=\:0$

$\displaystyle \text{Then: }\;m^2x \,-\,a \:=\:0 \qquad\Rightarrow\qquad x \:=\:\frac{a}{m^2}$


$\displaystyle \text{The line and the parabola intersect exactly }once.$
. . $\displaystyle \text{Therefore, the line is }tangent\text{ to the parabola.}$


Last edited by skipjack; February 28th, 2018 at 02:57 PM.
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December 4th, 2009, 06:59 PM   #4
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Re: tangent

thank you
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December 4th, 2009, 11:15 PM   #5
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For t$\neq$0, the tangent through the point (at², 2at) on the parabola has slope 1/t, so its equation is y = (1/t)x + at, i.e., y = mx + a/m, where m = 1/t. For m = 0, however, y = mx + a/m doesn't define a line, so it isn't the case that the given equation is that of a tangent to the parabola for all values of m.

Last edited by skipjack; February 28th, 2018 at 02:58 PM.
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December 5th, 2009, 01:13 AM   #6
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Re:

Quote:
Originally Posted by skipjack View Post
For t$\neq$0, the tangent through the point (at², 2at) on the parabola has slope 1/t, so its equation is y = (1/t)x + at, i.e., y = mx + a/m, where m = 1/t. For m = 0, however, y = mx + a/m doesn't define a line, so it isn't the case that the given equation is that of a tangent to the parabola for all values of m.

True, I didn't realize that until you told me. So it would be valid for all m, except for m=0.

Last edited by skipjack; August 20th, 2019 at 04:13 AM.
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December 5th, 2009, 03:01 PM   #7
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Correct.
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