# Thread: Tangent & Normal Q11

1. ## Tangent & Normal Q11

The equation of a curve is $\displaystyle y=\frac{2x}{3x+1}$. The normal to the curve meets the curve at point A(-1,1). Find the equation of the normal.

I am aware that I have to solve this question using the formula $\displaystyle (y-y_1)=\frac{1}{m}(x-x_1)$.

However, I am facing problems in simplifying the equation... I would be grateful if someone would show a detailed workings on it, thanks!

2. Originally Posted by Punch
The equation of a curve is $\displaystyle y=f(x)=\frac{2x}{3x+1}$. The normal to the curve meets the curve at point A(-1,1). Find the equation of the normal.

I am aware that I have to solve this question using the formula $\displaystyle (y-y_1)=\frac{1}{m}(x-x_1)$.

However, I am facing problems in simplifying the equation... I would be grateful if someone would show a detailed workings on it, thanks!
1. Let $\displaystyle P\left(p,\ \frac{2p}{3p+1} \right)$ denote the point on the curve whose normal you want to determine.

2. The slope of the normal is $\displaystyle m = -\frac1{f'(p)}$. To calculate f'(x) use the quotient rule:

$\displaystyle f'(x)=\frac2{(3x+1)^2}$ ..... $\displaystyle \implies$ ..... ..... $\displaystyle f'(p)=\frac2{(3p+1)^2}$ $\displaystyle \implies$ ..... $\displaystyle m = -\tfrac12 (3p+1)^2$

3. The slope of the normal is equal to the slope of PA:

$\displaystyle m = \dfrac{\frac{2p}{3p+1} - 1}{p+1}$

4. Both m-values must be equal:

$\displaystyle \dfrac{\frac{2p}{3p+1} - 1}{p+1} = -\tfrac12 (3p+1)^2$

Solve for p. Afterwards determine the equation of the normal.