Equation of normal to a curve given parametrically

Question

I was given a curve parametrically: $x = \frac{t}{(t^2)+1}$ and $y=\frac{t^2}{(t^2)+1}$ and was asked to find the equation of the normal to the curve at the point $t = \frac{1}{2}$.

I used parametric differentiation to work out the gradient which I worked out to be $\frac{2t}{1-t^2}$ and then substituted $t = \frac{1}{2}$ in $x, y$ and $\frac{dy}{dx}$ which resulted in $\frac{dy}{dx}=\frac{4}{3}$, $x =\frac{2}{5}$ and $y = \frac{1}{5}$. Then I tried using the equation $y-y_1 = m(x-x_1)$, i.e. $y-\frac{1}{5}=\frac{4}{3}(x-\frac{2}{5})$ which simplifies into $3y=4x+1$.

However, the correct answer is $4y=-3x+2$. Can somebody point out where I went wrong?

Answer 1

What you found is the slope of the tangent to the curve.

$m = \cfrac{dy}{dt} / \cfrac{dx}{dt} = \frac{2t}{1-t^2}$

Then the slope of the normal to the curve is,

$ = - \cfrac{1}{m} = \cfrac{t^2 - 1}{2t}$

At $t = \frac{1}{2}, $ the slope of the normal line is $ - \frac{3}{4}$

You already found $x, y$ coordinates at $t = \frac{1}{2}$. Completing the working gives the correct equation of the normal to the curve, as mentioned in your question.

Answer 2

The curve, in Cartesian coordinates, is a circle:

$x^2+y^2-y=0$,

centered at the point $C(0,\frac{1}{2})$ and radius $r=\frac{1}{2}$.

Equalizing $t^{2}+1$, in the expressions of $x$ and $y$, we get that $t=\frac{y}{x}$; replacing $t$ in one of the two coordinates, we get the circumference equation.

The point on the curve in $t=\frac{1}{2}$, is $(x,y)=(\frac{2}{5},\frac{1}{5})$.

The normal to the circumference at the point $(\frac{2}{5},\frac{1}{5})$, is none other than the line passing through it and the center of the circumference: $\frac{y-\frac{1}{2}}{\frac{1}{5}-\frac{1}{2}}=\frac{x-0}{\frac{2}{5}-0}$,

$y=\frac{-3x}{4}+\frac{1}{2}$.