# Basic Implicit Differentiation

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 $t=\pm 4$.

• Where does the "y - t = ..." come from? May 14, 2016 at 7:33
• @Arjun: The equation of the line whose gradient is $a$ passing through $(b,c)$ is given by $y-c=a(x-b)$. May 14, 2016 at 7:36
• Now that I think about it, that does make sense. I never learned that formally, and was only used to using $y = mx+c$. Thanks a lot for your answer! May 14, 2016 at 7:40