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 $(1, 0)$.
Here's what I did:
And,
$m_{normal} = \frac{y}{2x}$
But then here's where I get stuck - when I substitute $0$ into $y$ (in $y = mx + c$), I get the gradient and x term cancelled out, leaving me with $c = 0$.
How can I proceed to get the solutions?
Maybe you think that $(1,0)$ is on the curve $2x^2+y^2=18$.
$(1,0)$ is not on the curve, so the gradient is not zero.
Let $(s,t)$ be the point on the curve.
From what you wrote, the equation of the normal is given by $$y-t=\frac{t}{2s}(x-s)$$ Since this passes through $(1,0)$, we get $$0-t=\frac{t}{2s}(1-s),$$ i.e. $$t(s+1)=0$$
With $2s^2+t^2=18$, if $t=0$, then $s=\pm 3$, and if $s=-1$, then $t=\pm 4$.