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I would like to understand the concept of deriving an equation, given values. E.g. Derive the equation of parabola whose vertex is at origin and focus $(-3,0)$.

From this, I reckon the ends are curving left with the focus at $(-3,0)$. I am aware the distance from the focus to the vertex is equal to distance from vertex to directrix.

I also think (not sure) that with this, the equation would be $$y^2 = 4px$$ $$y^2 = 4\times -3\times x$$ Can I conclude, \therefore that the equation is $y^2 = -12x$ ?


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up vote 1 down vote accepted

I am assuming you are trying to find the parabola as the locus of points equidistant from focus and directrix.

The focus you are given at $F=(-3,0)$.

You are not given the directrix, but the vertex at $V=(0,0)$. Now the directrix is a line which is perpendicular to the line $FV$, on the opposite side of $V$ from $F$ and at the same distance from $V$ as $F$ is. [That's what the vertex is, plus the symmetry of the parabola]. So the directrix is the line $x=3$.

Now consider the point $(x,y)$ on the parabola. The distance from the line $x=3$ is $|(x-3)|$, and the distance from the point $(-3,0)$ is $\sqrt{(x+3)^2+y^2}$

Squaring the distances and equating the squares we find $(x-3)^2=(x+3)^2+y^2$ which simplifies to $y^2=-12x$.

The negative sign is because the directrix is to the right of the focus and reflects the orientation of the parabola.

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am wondering why you said "The distance from the line x=3 is (x−3) AND NOT (y-0) ...(x,y) and (3,0) – Sylvester Sep 11 '13 at 5:10
@Sylvester The distance between a point and a line is the length of the perpendicular from the point to the line. The line $x=3$ is vertical, so the distance will be measured in the horizontal direction. Why $-3$? Well if you can't see it, try points on the $x=axis$ - I've put an absolute value formula in my answer now - the sign of $z-3$ only keeps track of which side of the line the point is, but disappears when you square the distance. – Mark Bennet Sep 11 '13 at 6:10

As far as I know and by considering the coordinates of the focus $F(-3,0)$, the equation of parabola is: $$y^2=-2px$$ wherein $F(-p/2,0)$. So, here, $-p/2=-3$ and then $p=6$ and so $y^2=-12x$ would be the result.

enter image description here

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am a bit lost. The books am studying seem to mention that the equations of the parabola are x^2 = 4py and y^2=4px. – Sylvester Sep 10 '13 at 19:55
@BabakS.: I love plotting too, nice picture to go along with the answer! +1 – Amzoti Sep 10 '13 at 21:17
I agree with Amzoti! +1 – amWhy Sep 11 '13 at 0:52

If you mean by "derive" how the equation is obtained:

A parabola is the set of points "equidistant" from a line and a point external to that line, where "distance from a line" is taken to be the perpendicular distance. So you want the points $ \ (x,y) \ $ for which

$$ \sqrt{ ( x - [-3])^2 + ( y - 0 )^2} \ = \ \sqrt{(x-3)^2} \ , $$

since the directrix is the vertical line $ \ x = 3 \ $ ; the left-hand side represents the distance from $ \ ( x,y ) \ $ to the focus $ \ (-3,0) \ $ and the right-hand side gives the perpendicular distance from $ \ ( x, y ) \ $ to the directrix, which would be measured along a horizontal line intersecting the directrix at $ \ (3, y) \ $ .

We can square both sides to obtain

$$ (x+3)^2 + y^2 \ = \ (x-3)^2 \ \Rightarrow \ x^2 + 6x + 9 + y^2 \ = \ x^2 - 6x + 9 \ $$

$$\Rightarrow \ 6x + y^2 \ = \ - 6x \ \Rightarrow \ y^2 = -12x \ , $$

a "horizontal" parabola opening "to the left". This argument can be generalized pretty easily to parabolas opening vertically as well.

The parabola is the "easy" conic section to derive; the ellipse and hyperbola require somewhat more effort...

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