# Differentiation question find the normal to the curve

The curve $C$ has equation $2x^2+y^2=18$. Determine the coordinates of the four points on $C$ at which the normal passes through the point $(1,0)$.

I got the gradient of the normal to be: (I called the co-ordinates $a,b$)

$y = (b/2a)x+b/2$,

then I don't know how to continue, note i found this by differentiating,

thanks

Differentiate explicitly:

$$4x+2yy'=0\implies y'=-\frac{2x}y$$

So we need points $\;(x_0,y_0)\;$ on the curve s.t. that their normal:

$$y-y_0=\frac{y_0}{2x_0}(x-x_0)$$

passes through $\;(1,0)\;$ , meaning

$$-y_0=\frac{y_0}{2x_0}(1-x_0)\iff-2x_0y_0=y_0-x_0y_0\iff-x_0y_0=y_0$$

1) If $\;y_0=0\;$ then $\;2x_0^2=18\implies x_0=\pm3\;$

2) If $\;y_0\neq0\;$ then $\;x_0=-1\;$ ...and now you continue.