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 December 4th, 2009, 08:42 AM #1 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 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.
 December 4th, 2009, 10:36 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 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.
December 4th, 2009, 12:59 PM   #3
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Re: tangent

Hello, mikeportnoy!

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

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.

 December 4th, 2009, 06:59 PM #4 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 Re: tangent thank you
 December 4th, 2009, 11:15 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,119 Thanks: 2331 For t$\neq$0, the tangent through the point (atÂ², 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.
December 5th, 2009, 01:13 AM   #6
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Re:

Quote:
 Originally Posted by skipjack For t$\neq$0, the tangent through the point (atÂ², 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.

 December 5th, 2009, 03:01 PM #7 Global Moderator   Joined: Dec 2006 Posts: 21,119 Thanks: 2331 Correct.

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