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I'm sorry for the stupid question, but it seems that extensive googling didn't yield an answer.

I've learned about parabolas, and how the parabola is the curve that is equidistant from a point (Focus) and a line that is perpendicular to the axis of the parabola.

But then, while studying classical mechanics, you come upon the fact that a projectile follows a parabolic path! And it seems that the only justification i can find for it is that since the path of the projectile is given by a quadratic function, it's curve is a parabola.

However, i can't seem to find a proof that the curve of quadratic function is the curve given by the set of points equidistant from a particular point and a line.

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I guess you are asking the following question --- “Why is that ALL the quadratic curves are in the shape of a parabola?”

First of all, we have to agree that the graph of the basic quadratic function $y = f(x) = x^2$ is a parabola. This can be verified by simple plotting.

Secondly, any quadratic function can be re-written into the standard form---$y = ax^2 + bx + c$ for some $a$, $b$ and $c$ with $a$ not equal to zero.

By completing square, that standard form can be further re-written as $y = a(x – h)^2 + k$ for some $h$ and $k$.

The last equation can be interpreted as:-

(1) perform a right shift of the graph of $y = f(x)$ $h$ units (or left if $h$ is negative); followed by

(2) amplifying the graph obtained in (1) by $|a|$ units (or shrinking depending on the value of $a$). If $a$ is negative, the graph is further flipped upside down about the x-axis; followed by

(3) perform an up/down shift by $k$ units (depending on the value of $k$) to the resultant graph obtained in (2)

The translation actions in (1) and (3) as well as the flipping action in (2) are just moving the graph around and they did not change the shape of the original.

The amplification action will cause the opening of the graph narrower or wider but did not change the shape of original.

That is why all the quadratic curve are parabola in shape.

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  • $\begingroup$ Yes, but doesn't simple plotting not necessarily prove that the curve is a parabola? Unlike a square or a rectangle, we can only say that the curve looks like a parabola. $\endgroup$ – Aayush Agrawal Mar 6 '16 at 3:57
  • $\begingroup$ @AayushAgrawal Note that all we need to plot is one graph (of $y = f(x) = x^2$, the basic one) and verify that it is indeed a parabola. As shown in my post, all other quadratic graphs are just images of the basic under suitable successive transformations. $\endgroup$ – Mick Mar 6 '16 at 12:32
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Given a point $(a,b)$ and a horizontal line $y=k$, in $\mathbb{R^2}$.

Let the locus of the points which are equally far away from the point and the line be denoted by $(x,y)$.

Then distance between $(x,y)$ and the line is just $|y-k|$

The distance between the point $(x,y)$ and $(a,b)$ is $\sqrt{(x-a)^2 + (y-b)^2}$

Equating the two:

$|y-k|=\sqrt{(x-a)^2 + (y-b)^2}$

$(y-k)^2=(x-a)^2+(y-b)^2$

Expanding and rearranging: $y=\frac{x^2-2ax+(a^2+b^2-k^2)}{2b-2k}$

Given any quadratic function, you can find the unique value for $a,b,k$ (thus the diretrix and focus).

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\begin{align} & \text{distance from } (x,x^2) \text{ to } \left( 0, \frac 1 4 \right) \\[10pt] = {} & \sqrt{(x-0)^2 + \left(x^2 - \frac 1 4 \right)^2} = \sqrt{x^2 + \left(x^4 -\frac 1 2 x^2 + \frac 1 {16}\right)} \\[10pt] = {} & \sqrt{x^4 + \frac 1 2 x^2 + \frac 1 {16}} = \sqrt{\left( x^2 + \frac 1 4 \right)^2} \\[10pt] = {} & x^2 + \frac 1 4 = \text{distance from }(x,x^2) \text{ to the line } y = - \frac 1 4. \end{align}

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