Differentiation question find the normal to the curve Hi I was struggling to this question, can anyone please help me :P
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
 A: 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.
A: *

*Differentiate and get slope of tangent (say m)

*Then slope of normal = -1/(slope of tangent) = -1/m
Now you have the required slope (in form of x,y) and a point from which it should pass

*

*Use point slope form of straight lines (y-y1)=m(x-x1) where x1,y1 are the required point from which the straight line should pass and m is the required slope. NOTE here m is the slope of normal which was found in 2nd step


*Now you will get some expression in x,y.. get the values of x and y from it and put them in the given equation of curve C
Also its an ellipse.. look closely you will see it's along the y-axis so you can directly get two required points by the points which cut the x-axis.. and also the other points will be symmetrical along the x-axis (say you get x=-1,y=4 then x=-1,y=-4 will also be the required point) Only in this case
So by looking closely you can directly get the answer :)
