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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 nevertheless. 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: 5  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 , and point T has coordinates , then the slope of the secant line is and you want a point where the tangent line has the same slope: solve for x. 
September 22nd, 2011, 01:20 PM  #3 
Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 132 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus  Re: Point at which tangent line is parallel to secant? [color=#000000]Look at the figure.[/color] [attachment=0:2i07jacw]sec.png[/attachment:2i07jacw] 
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 $\displaystyle slope=\frac{y_py_t}{x_px_t}$. 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 $\displaystyle T(2,4)$ $\displaystyle slope=\frac{04}{02}=\frac{4}{2}=2$ $\displaystyle f'(x)=2x$ $\displaystyle f'(x)=slope$ $\displaystyle 2x=2$ 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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