Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$ Determine the point of intersection between $f(x) = x^2$ and its normal in the point $(a, a^2)$
Answer:
This should be easy enough...
$f'(x) = 2x$
The tangent line in the point $(a, a^2)$ is $y - a^2 = 2 (x - a) \rightarrow y = 2x + a^2 - 2a$
The equation for the normal line is: $y - a^2 = -\frac{1}{2}(x - a) \rightarrow y = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
Now to determine the point of intersection we just see when $f(x)$ and the normal line is equal, i.e.
$x^2 = -\frac{1}{2}x + a^2 + \frac{1}{2}a$
But this seems like a nonsense equation...

The slope of the normal line is going to be: $-\frac{1}{2a}$.
set $g(x)=-\frac{1}{2a}x+(a^2+\frac{1}{2})$
You want to solve $g(x)=f(x)$.
$-\frac{1}{2a}x+(a^2+\frac{1}{2})=x^2$
$x^2+\frac{1}{2a}x+\frac{1}{16a^2}=(a^2+\frac{1}{2})+\frac{1}{16a^2}$
$x=\pm\sqrt{(a^2+\frac{1}{2})+\frac{1}{16a^2}}-\frac{1}{4a}$
Personally, I prefer vectors, but it amounts to the same thing:
$(a,a^2)+t[-2a,1]=(x,x^2)$
$a^2+t=(a(1-2t))^2$
And you can solve in this way to get a $t$, which I think is more intuitive, and less "symbol pushy"
