Find, in terms of $s$,the coordinates of the point where this normal cuts the curve again.

Question

a) Find the equation of the normal at the point $(2s,\frac{2}{s})$ to the curve whose parametric equations are $x=2s,y=\frac{2}{s}$

b) Find, in terms of $s$,the coordinates of the point where this normal cuts the curve again.

I didn't have any problems with the part a) using implicit differentiation and the fact that $y-y_1=m(x-x_1)$ to get the equation of the normal:

$y-\frac{2}{s}=s^2(x-2s)$

which is correct according to the answers at the back of the book. I also know that the normal will cut the curve again at a point which satisfies the equation of the normal and $x=2s$ and $y=\frac{2}{s}$

I'm just having a bit of problem substituting these values of x and y into the equation of the normal and basically getting zero. I must be going about this the wrong way here - the textbook gives the answer as $(-\frac{2}{s^3},-2s^3)$

Any hints/explanations much appreciated.

Answer 1

The parametric equation of the curve is $x=2s$ & $y=\frac{2}{s}$

Now, the equation of the curve in cartesian coordinates is obtained by eliminating $s$ as follows $$y=\frac{2}{\frac{x}{2}}=\frac{4}{x}$$ $$\implies \color{blue}{xy=4}$$ Above, equation represents a rectangular hyperbola.

Now, differentiating the equation w.r.t. $x$ to obtain the slope of tangent
$$\frac{d}{dx}(xy)=\frac{d}{dx}(4)$$ $$x\frac{dy}{dx}+y\frac{d}{dx}(x)=0$$ $$\frac{dy}{dx}=\frac{-y}{x}$$ Hence, the slope of normal at the point $\left(2s, \frac{2}{s}\right)$ $$=\left[\frac{-1}{\frac{dy}{dx}}\right]_{x=2s, y=\frac{2}{s}}=\left[\frac{-1}{\frac{-y}{x}}\right]_{x=2s, y=\frac{2}{s}}=\frac{2s}{\frac{2}{s}}=s^2$$ Hence, the equation of the normal pasing through the point $\left(2s, \frac{2}{s}\right)$ & having slope $m=s^2$ $$y-\frac{2}{s}=s^2(x-2s)$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{s^3x-sy+2(1-s^4)=0}}$$ Now, solving the equations of the curve & the normal as follows $$s^3x-s\frac{4}{x}+2(1-s^4)=0$$ $$s^3x^2+2(1-s^4)x-4s=0$$ Now, solving above quadratic equation for $x$ as follows $$x=\frac{-2(1-s^4)\pm\sqrt{4(1-s^4)^2-4(s^3)(-4s)}}{2s^3}$$ $$x=\frac{-2(1-s^4)\pm 2(1+s^4)}{2s^3}\iff x=2s, x=\frac{-2}{s^3}$$ Now, substituting the values of $x$ we get corresponding y-coordinates $$x=2s\implies y=\frac{4}{2s}=\frac{2}{s}$$ $$x=\frac{-2}{s^3}\implies y=\frac{4}{\frac{-2}{s^3}}=-2s^3$$ Hence the points of intersection of normal & curve are $\left(2s, \frac{2}{s}\right)$ & $\left( \frac{-2}{s^3}, -2s^3\right)$

Hence it is clear that the normal at point $\left(2s, \frac{2}{s}\right)$ cuts the curve at another point given as
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\left(\frac{-2}{s^3}, -2s^3\right)}}$$

Answer 2

Let the normal intersect with the curve again at the point $(2t,\frac 2t)$ and substitute this for $(x,y)$ in the equation of the normal. You then have to look for the common factor $(s-t)$ in the equation $$\frac 2t-\frac 2s=s^2(2t-2s)$$ which can be cancelled since $t\neq s$, which gives you $t=-\frac{1}{s^3}$ and hence the answer.