Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $y^2 = -12x$ ?

Thanks.

share|improve this question

3 Answers 3

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.

share|improve this answer
    
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

share|improve this answer
    
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...

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.