My Math Forum Point at which tangent line is parallel to secant?

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 September 9th, 2011, 07:07 AM #1 Member   Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 Point at which tangent line is parallel to secant? Got this question which is probably a no brainer, but confusing to me never-the-less. How do I find the the point at which the tangent line is parallel to the secant PT Picture of graph:
 September 10th, 2011, 03:55 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Point at which tangent line is parallel to secant? Well, that would depend upon exactly how the function is given. If that is a graph of y= f(x), Point P has coordinates $$$x_P,y_P$$$, and point T has coordinates $$$x_T,y_T$$$, then the slope of the secant line is $\frac{y_P-y_T}{x_P-x_T}$ and you want a point where the tangent line has the same slope: solve $f'(x)= \frac{y_P-y_T}{x_P-x_T}$ for x.
September 22nd, 2011, 01:20 PM   #3
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Re: Point at which tangent line is parallel to secant?

[color=#000000]Look at the figure.[/color]

[attachment=0:2i07jacw]sec.png[/attachment:2i07jacw]
Attached Images
 sec.png (38.7 KB, 467 views)

 May 11th, 2017, 11:26 PM #4 Newbie   Joined: May 2017 From: Pondicherry Posts: 1 Thanks: 0 then what can we say about the points P,T@R?? are they following any sequence like A.P, G.P etc.,?
 May 12th, 2017, 09:47 AM #5 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 Calculate the slope of the secant line via $\displaystyle slope=\frac{y_p-y_t}{x_p-x_t}$. Then if you take the derivative of the function and make it equal to slope of the secant. $\displaystyle f'(x)=slope$ Solve for x. Put x back into the original function and solve for y. That is the point on the function that the tangent is parallel to the secant $\displaystyle \overline{PT}$ Practical Example: $\displaystyle f(x)=x^2$ $\displaystyle P(0,0)$ and $\displaystyle T(2,4)$ $\displaystyle slope=\frac{0-4}{0-2}=\frac{-4}{-2}=2$ $\displaystyle f'(x)=2x$ $\displaystyle f'(x)=slope$ $\displaystyle 2x=2$ so $\displaystyle x=1$ $\displaystyle f(1)=1$ So at point $\displaystyle (1,1)$ on the function $\displaystyle f(x)=x^2$, the line tangent to the function is parallel at $\displaystyle (1,1)$ to the line secant at points $\displaystyle (0,0)$ and $\displaystyle (2,4)$ There is also the possibility they just want you to eyeball it, if they haven't given you all the information, such as f(x) and such. In this case, ZardoZ hooked you up with how you'd eyeball it, by drawing a line between $\displaystyle \overline{PT}$ and finding a point that the line tangent would be a match to the slope of the secant. Hope this helps Last edited by phrack999; May 12th, 2017 at 09:50 AM.

 Tags line, parallel, point, secant, tangent

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