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July 5th, 2009, 04:50 PM   #1
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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.
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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 $\displaystyle \frac{1}{2\sqrt{2}}(x-2)+2\sqrt{2}$.
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Last edited by skipjack; February 28th, 2018 at 01:39 PM.
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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.
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July 5th, 2009, 06:11 PM   #4
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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.
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July 5th, 2009, 06:40 PM   #5
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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.
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July 5th, 2009, 06:43 PM   #6
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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.
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July 6th, 2009, 03:31 AM   #7
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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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