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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 (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. 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 (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. December 5th, 2009, 03:01 PM #7 Global Moderator   Joined: Dec 2006 Posts: 21,119 Thanks: 2331 Correct. Tags tangent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post johnny Trigonometry 4 March 1st, 2011 08:31 AM jordanshaw Trigonometry 4 October 12th, 2010 09:41 PM jens Calculus 2 November 2nd, 2008 01:16 PM me Trigonometry 3 January 31st, 2008 10:30 AM jordanshaw Calculus 4 December 31st, 1969 04:00 PM

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