User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 July 4th, 2015, 12:28 PM #1 Newbie   Joined: Mar 2015 From: uk Posts: 13 Thanks: 0 finding equation tangent to an equation If I have an equation y =f(x) and I know the coords (x1,y1) of a point that does not lie on the line y =f(x). If I draw a straight line from (x1,y1) then at what point on f(x) will the staright line be perpendicular to a tangent of f(x) and hence whats the equation of this tangent in terms of the original point x1,y1? Thank you July 4th, 2015, 01:03 PM   #2
Senior Member

Joined: Feb 2010

Posts: 714
Thanks: 151

Quote:
 Originally Posted by DarkLink94 If I have an equation y =f(x) and I know the coords (x1,y1) of a point that does not lie on the line y =f(x). If I draw a straight line from (x1,y1) then at what point on f(x) will the staright line be perpendicular to a tangent of f(x) and hence whats the equation of this tangent in terms of the original point x1,y1? Thank you
You state that $\displaystyle (x_1,y_1)$ does not lie on the line $\displaystyle y=f(x)$.

Lines have no tangents. So, as stated, your problem makes no sense. Think about it and re-phrase it. July 4th, 2015, 01:39 PM #3 Newbie   Joined: Mar 2015 From: uk Posts: 13 Thanks: 0 the equation f(x) has tangents. You've understood what I said wrong. The point (x1,y1) does not indeed lie on f(x). Im wondering this... if I draw a straight line from (x1,y1), at what points does it intersect a tangent of f(x) that is also perpendicular to the line from (x1,y1). 1) the line intersects f(x) 2) at the point that the tangent to f(x) is perpendicular to the line from (x1,y1) Last edited by DarkLink94; July 4th, 2015 at 01:47 PM. July 4th, 2015, 02:15 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Yes, he understood what you meant to say, but that's not what you said! First, you said "the line y= f(x)". If f is not a linear function of x, that is a curve not a line. If f is linear, then it is a line but lines do not have "tangents" (unless you consider the line its own "tangent"). If y= f(x) is a general curve then there may be a number of places at which a line through $\displaystyle (x_1, y_1)$ is perpendicular to the curve. For example, if $\displaystyle y= \sqrt{1- x^2}$, and $\displaystyle (x_1, y_1)= (0, 0)$, then every line through (0, 0) is perpendicular to a tangent of y= f(x). You can try to do this: Let $\displaystyle (x_0, y_0)$ be the point on the graph of y= f(x) at which the line through $\displaystyle (x_1, y_1)$. The slope of that line is $\displaystyle \frac{y_1- y_0}{x_1- x_0}$ while the slope of the line perpendicular to the graph at $\displaystyle (x_0, y_0)$ is $\displaystyle -\frac{1}{f'(x_0)}$. So we must have $\displaystyle \frac{y_1- y_0}{x_1- x_0}= -\frac{1}{f'(x_0}$. That is a single equation with two unknowns, $\displaystyle x_0$ and $\displaystyle y_0$ so there may be many solutions. July 4th, 2015, 02:19 PM   #5
Senior Member

Joined: Feb 2010

Posts: 714
Thanks: 151

Quote:
 Originally Posted by DarkLink94 You've understood what I said wrong.
You said "the line $\displaystyle y=f(x)$". So I did understand what you said. You just didn't say what you meant to say ... namely "the curve $\displaystyle y=f(x)$".

I now think I understand what you want.

You are given a curve $\displaystyle y=f(x)$ and a point $\displaystyle (x_1,y_1)$ not on the curve. What you want are the coordinates $\displaystyle (x_2,y_2)$ of a point on the curve so that the line through the two points is perpendicular to the tangent at $\displaystyle (x_2,y_2)$. Solve these two equations simultaneously:

$\displaystyle y_2=f(x_2)$

$\displaystyle f^{ \prime}(x_2) = \dfrac{x_1-x_2}{y_2-y_1}$ July 4th, 2015, 02:32 PM #6 Newbie   Joined: Mar 2015 From: uk Posts: 13 Thanks: 0 thank you very much- exactly what I wanted. Yes I did say f(x) was a line... obviously wrong- apologies. Tags equaion, equation, finding, tangent 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 Shamieh Calculus 3 October 9th, 2013 05:32 AM Shamieh Calculus 2 October 3rd, 2013 07:17 AM jezb5 Calculus 2 January 31st, 2013 02:22 AM missfuzzy Trigonometry 7 May 14th, 2009 05:52 PM jezb5 Trigonometry 0 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      