Differentiation(Finding coordinates of a point on the curve)

[wei1006]

Homework Statement

Homework Equations

(y-y1)/(x-x1)=m

The Attempt at a Solution

I have attempted part i but I don't know how to do part ii. As point B is still part of the curve and the normal, do I still sub with the same normal eqn? :/ I have no idea how to start... Please help thanks[/B]

 

[LCKurtz]

Please post a right-side up image.

 

[HallsofIvy]

For those who don't want to stand on their heads, the problem says,
"The normal to the graph of $y= \frac{2x}{3x+ 1}$ at (-1, 1) meets the curve again at point B. Find
(i) the equation of the normal.
(ii) the coordinates of B."

If the problem is not important enough to you to simply type that in, why would it be important enough for us to try to read and solve it?

The derivative of $\frac{2x}{3x+ 1}$, at any point, (x, y), is $\frac{2(3x+ 1)- 3(2x)}{(3x+ 1)^2}= \frac{2}{(3x+ 1)^2}$. At (-1, 1) that is $\frac{1}{2}$ so the slope of the normal line is -2. That part you have right. However, if you put x= -1 in the equation you give, $y= -2x+ \frac{3}{2}$, you get $y= -2(-1)+ \frac{3}{2}= \frac{7}{2}$, NOT 1. You seem to have the idea that you find the equation of the normal by finding the equation of the tangent line, then replacing the slope, m, with -1/m. That is NOT correct. The slope of the normal line is -1/m but the constant term, "b" in "y= mx+ b", also has to be changed to give the correct point.

With $y= -2x+ b$, to get y= 1 when x= -1, you must have $1= 2+ b$ so b= -1. The equation of the normal line is $y= -2x- 1$.

That line will cross $y= \frac{2x}{3x+ 1}$ when $-2x+ 1= \frac{2x}{3x+ 1}$. If you multiply both sides of that equation by the denominator on the right, you get a quadratic equation. The line clearly crosses the graph of $y= \frac{2x}{3x+ 1}$ in two places with the x value of those two points the two roots of the quadratic equation. One solution is the given point, (1, -1), the other is B.

 

[wei1006]

I see, thank you for the detailed explanation! I apologise for not adjusting the angle of the picture posted and not able to type out the question(it will only look messy and confusing when I type out the equation)as I have no access to the computer in school and was using a phone at that point of time. The question is definitely important to me and I really appreciate your efforts in helping me with clearing this misconception. Sorry for the inconvenience caused and thank you once again.