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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.

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