Hi I was struggling to this question, can anyone please help me 😛
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 $a,b$)
$y$ = $\dfrac{b}{2a} \ x+\dfrac{b}{2}$,
then I don't know how to continue, note i found this by differentiating,
thanks
Best Answer
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.