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 July 5th, 2009, 04:50 PM #1 Member   Joined: Jul 2009 Posts: 40 Thanks: 0 Write the equation of the tangent line by (2,2√2) I already graphed x^2-6x+y^2=0, which gave me a circle , but how in the world should I write an equation? I think it might be a linear equation. Any step on how to solve? Thanks in advance. Last edited by skipjack; February 28th, 2018 at 01:34 PM. July 5th, 2009, 05:20 PM   #2
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re: Write the equation of the tangent line by (2,2√2)

I'm assuming you know calculus.
Use implict differentiation to find the derivative of the equation:
$\displaystyle 2x-6+2y\frac{dy}{dx}=0$
$\displaystyle 2y\frac{dy}{dx}=-2x+6$
$\displaystyle \frac{dy}{dx}=\frac{-x+3}{y}$
Then find the slope of the tangent line at (2, 2√2):
$\displaystyle \frac{-x+3}{y}$
$\displaystyle \frac{1}{2\sqrt{2}}$
You can find the equation of the tangent line by using point-slope form:
$\displaystyle \frac{1}{2\sqrt{2}}(x-2)+2\sqrt{2}$

So the equation of the tangent line is $\displaystyle \frac{1}{2\sqrt{2}}(x-2)+2\sqrt{2}$.
Attached Images wztangenteq2img.gif (20.1 KB, 97 views)

Last edited by skipjack; February 28th, 2018 at 01:39 PM. July 5th, 2009, 06:07 PM   #3
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re: Write the equation of the tangent line by (2,2√2)

Quote:
 Originally Posted by SidT I'm assuming you know calculus. Use implict differentiation to find the derivative of the equation: $\displaystyle 2x-6+2y\frac{dy}{dx}=0$ $\displaystyle 2y\frac{dy}{dx}=-2x+6$ $\displaystyle \frac{dy}{dx}=\frac{-x+3}{y}$ Then find the slope of the tangent line at (2, 2√2): $\displaystyle \frac{-x+3}{y}$ $\displaystyle \frac{1}{2\sqrt{2}}$ You can find the equation of the tangent line by using point-slope form: $\displaystyle \frac{1}{2\sqrt{2}}(x-2)+2\sqrt{2}$ So the equation of the tangent line is $\displaystyle \frac{1}{2\sqrt{2}}(x-2)+2\sqrt{2}$.
Brilliantly presented.

Last edited by skipjack; February 28th, 2018 at 01:37 PM. July 5th, 2009, 06:11 PM #4 Member   Joined: Jul 2009 Posts: 40 Thanks: 0 re: Write the equation of the tangent line by (2,2√2) Thank you so much and by the way no I have not taken Calculus yet; that is why I had no idea on solving my question. Thank you very much. Last edited by skipjack; February 28th, 2018 at 01:31 PM. July 5th, 2009, 06:40 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,941 Thanks: 2210 Given the point (x_0, y_0) on the circle with equation $\displaystyle (x\,-\,a)^{\small2}\,+\,(y\,-\,b)^{\small2}\,=\,r^{\small2},$ the tangent through that point has equation $\displaystyle (x_0\,-\,a)(x-a)\,+\,(y_0\,-\,b)(y\,-\,b)\,=\,r^{\small2},$ which is easy to remember. Last edited by skipjack; February 28th, 2018 at 01:31 PM. July 5th, 2009, 06:43 PM #6 Member   Joined: Jul 2009 Posts: 40 Thanks: 0 re: Write the equation of the tangent line by (2,2√2) hmmm good point, that is easier. Also, to write an equation to the circle symmetric about the x axis to the line. Would it be this? 1/2√2(x-2)-2√2. It seems right to be since the original tangent line is 1/2√2(x-2)+2√2. Last edited by skipjack; February 28th, 2018 at 01:36 PM. July 6th, 2009, 03:31 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,941 Thanks: 2210 No. To reflect any function in the x-axis, just negate the entire function (i.e., the graph of y = -f(x) is the reflection in the x-axis of the graph of y = f(x)). By the way, you should write your final answers as equations, as requested in the question. Tags 2√2, equation, line, lineby, tangent, write 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 unwisetome3 Calculus 2 October 28th, 2012 06:52 PM unwisetome3 Calculus 4 October 20th, 2012 07:38 AM kevpb Calculus 3 May 25th, 2012 10:32 PM arron1990 Calculus 5 February 9th, 2012 01:29 AM ArmiAldi Calculus 1 April 16th, 2008 05:51 AM

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