Prove that a tangent line at a point X of a parabola bisects the angle between the point vector and axis vector

Question

I'm trying to solve excercise $14.7.14$ in Tom Apostol's Calculus vol. $1$. I started with a parabola $y=\frac{cx^2}{2}$, where its focus is at point $(0, \frac{c}{2})$.

Then, expressed as a vector function of a real variable, the vector from the focus to the any point on the parabola is $X(x)=(x,\frac{c(x^2 - 1)}{2})$.

I would expect that the tangent line direction is $\tag{1} T(x) = (1, cx)$, and the axis line direction is $A=(0, 1)$.

I thought that I just need to prove that $\frac{XT}{\|X\|\|T\|} = \frac{TA}{\|T\|} = \frac{cx}{\sqrt{1 + c^2x^2}}$, because it would mean that the two angles are the same...but I don't get far with that.

Where am I going wrong?

Thanks!

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As pointed out in the answer, I had a typo in the focus location, as well as in the direction of the axis direction.

Actual focus is at $F=(0, \frac{1}{2c})$. The vector from focus to any point on the parabola is $X(x) = (x, \frac{c^2x^2 - 1}{2c})$. The tangent line at the parabola point is $T(x)=(1, cx)$. The axis line direction is $A = (0, 1)$.

Answer 1

It looks like you have the right idea, but there are several errors in your set-up.

1. If the equation of your parabola is $y = \frac{1}{2} cx^2$, the focus is at $\left( 0, \frac{1}{2c} \right)$, not $(0,c)$.
2. For what it's worth, when I read "point vector" I first thought you meant the vector from the origin/vertex to $X$, when you meant the vector from the focus to $X$. The coordinates you found for the latter are fine after correction (1), but I figured I'd point out that this is not standard terminology.
3. The axis line direction is any vector proportional to $\left( 0, 1 \right)$, not $(1, cx)$. Do you see why?