# Proof without calculus: slope of parabola

This is an image :D

For anyone who has learnt Calculus must know that the slope at any point on a parabola, for instance $y = \frac{1}{2} x^2$, is always equivalent to $\frac{dy}{dx} = x$. Yet I was just wondering if we can prove it without Calculus? Maybe with only geometry, vector, I don't know. If you would like to try this proof, I totally appreciate your effort, since I can't complete it on my own. Share your ideas here and let me know your wonderful thoughts, thanks! :D

For anyone who is curious to take a peek at this animation I made in Desmos(regarding this problem), feel free to check it out on this link: https://www.desmos.com/calculator/qw71a3cyyl

Note by Horace Hung
1 month ago

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The slope is equal to rise over run $\Big( \frac{\Delta y}{\Delta x } \Big)$. Consider two points separated by a very small $x$ distance $\epsilon$.

$(x_1, y_1) = \big(x_0, \frac{1}{2} x_0^2 \Big) \\ (x_2, y_2) = \big(x_0 + \epsilon, \frac{1}{2} (x_0 + \epsilon)^2 \Big)$

The slope is the change in $y$ over the change in $x$.

$\text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{1}{2} (x_0 + \epsilon)^2 - \frac{1}{2} x_0^2 }{(x_0 + \epsilon) - x_0}$

Expanding gives:

$\text{slope} = \frac{\frac{1}{2} x_0^2 + x_0 \epsilon + \frac{1}{2} \epsilon^2 - \frac{1}{2} x_0^2 }{\epsilon}$

Suppose $\epsilon$ is much smaller than $x_0$, meaning that the $\epsilon^2$ term in the numerator can be neglected. The expression then reduces to:

$\text{slope} = x_0$

- 1 month ago

That looks suspiciously like a derivative to me! :D

- 1 month ago

Isn't the very concept of "slope" the same thing as a derivative?

- 1 month ago

True. :)

I think that's what the original question is referring to though: is there a way to define/calculate slope in, say, a purely geometrical way? @Avinash Rathore seems to have described something like that.

- 1 month ago

Finally someone understands my question!!! @Steven Chase solution is in fact first principle in disguise, which I'd regard as part of Calculus :( Still a nice try. @Avinash Rathore solution is really similar to my original geometric proof, congrats!

Frankly I'd like to post my merely geometric proof here (without the slightest use of Calculus / Algebra / point-line distance formula), but the comment box is too small so it lefts as an exercise to readers

P.S. Just kidding, I'll try to post it somehow (Hint: @Avinash Rathore solution but with purely congruent triangles/locus of parabola/concepts of focus + directrix etc.)

- 1 month ago

Awesome! :D

- 1 month ago

The tangent and the normal at a point are the two angle bisectors of the lines (i) joining the focus and the point, and (ii) a line parallel to the y-axis passing through the point. Using the formula for the angle bisectors of a couple of lines,

          (Ax + By + C)^2/ (A^2 + B^2) = (ax + by + c)^2/ (a^2 + b^2)


(whose 2 roots give the tangent and normal), the line with the negative y intercept is the tangent. Simplifying, we can easily get the slope of the tangent as m = x.

NOTE - The above formula can be easily derived by equating the distance of any point on an angle bisector from the 2 given lines. This derivation does not require calculus.

Also, we note the above fact of the tangent and normal being angle bisectors by remembering the parabolic property that all lines parallel to the axis of the parabola get reflected or focused on to the focus which makes the normal at the point an angle bisector (and the tangent, being perpendicular to the normal is the other angle bisector).