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 September 27th, 2011, 12:22 PM #1 Member   Joined: Sep 2011 Posts: 49 Thanks: 0 Difficult parabola equation Hi I have an exercise and I’ve been struggling for a while with that. This is it: Find coordinates of the focus and the equation of the directrix of the parabola whose equation is: 3y"squer"=8x The chord which passes through the focus parallel to the directrix is called the latus rectum of the parabola. Show that the latus rectum of the above parabola has length8/3. I do not understand this exercise at all. I have always thought that parabola equation looks like that: y = 4x"squer" -4x +3 Anyone know how to resolve this problem?
 September 27th, 2011, 12:30 PM #2 Member   Joined: Aug 2009 Posts: 69 Thanks: 0 Re: Difficult parabola equation tip: It's not the common definition of a parabola. The equation you gave ($y= 4 x^{2}-4x+3$) is for a vertical parabola. The equation $3 y^{2}=8x$ is a horizontal parabola. Try using the equation $\frac{3}{8} y^{2}=x$, and treating that as a horizontal parabola.
 September 27th, 2011, 01:25 PM #3 Member   Joined: Sep 2011 Posts: 49 Thanks: 0 Re: Difficult parabola equation I know it’s a horizontal parabola. I’m using these formulas to search for focus: y2 = 2p • x (p/2,0) and x=-p/2 for directrix of the parabola, that gives me: y2 = 8/3x so x=4/3 and x=4/3 but the answer is: x=2/3 and -2/3.
 September 27th, 2011, 01:52 PM #4 Member   Joined: Aug 2009 Posts: 69 Thanks: 0 Re: Difficult parabola equation We know that in an equation of a horizontal parabola $x=ay^2+by+c$, or in our case where $b=c=0$, the focus is at (1/(4a),0), and $\frac{1}{4a}=\frac{2}{3}$. We want to find the two points where the given line intersects the parabola, and then find the length between them. Since the directrix of this horizonal parabola is vertical, we want to find a vertical line that passes through the focus. This is given by $x=\frac{2}{3}$. The two points of intersection are where $y^2=\frac{8}{3}\cdot \frac{2}{3}=\frac{4^2}{3^2}$, so the two solutions for y are $\frac{4}{3}$ and $-\frac{4}{3}$. The distance between these two points -- since our line is vertical -- is $\frac{8}{3}$.
 September 27th, 2011, 02:35 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,375 Thanks: 2010 If you wish to claim "we know" something that relates to a "standard" equation, it helps to mention what that equation is, so that it's clear where the constant "a" is to be found.
 September 27th, 2011, 02:46 PM #6 Member   Joined: Aug 2009 Posts: 69 Thanks: 0 Re: Difficult parabola equation changed.
 September 27th, 2011, 03:05 PM #7 Member   Joined: Sep 2011 Posts: 49 Thanks: 0 Re: Difficult parabola equation thanks NeuroFuzzy.
 September 27th, 2011, 03:40 PM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond Re: Difficult parabola equation $3y^2\,=\,8x\,\Rightarrow\,y^2\,=\,\frac{8}{3}x$ $4p\,=\,\frac{8}{3}\,\Rightarrow\,p\,=\,\frac{2}{3}$ $y^2\,=\,\frac{8}{3}\,\cdot\,\frac{2}{3}\,\Rightarr ow\,y\,=\,\pm\frac{4}{3}$ $\text{Focus: }$$\frac{2}{3},\,0$$\text{ Directrix: }x\,=\,-\frac{2}{3}\text{ Length of latus rectum: }\frac{8}{3}$
September 27th, 2011, 07:40 PM   #9
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Quote:
 Originally Posted by NeuroFuzzy changed.
Changed, but now nonsense.
For the standard equation $x\,=\,ay^2\,+\,by\,+\,c,$ the focus is at $$$\frac{1}{{\small4}a}\,-\,\frac{1}{{\small4}a}(b^{\small2}\,-\,4ac),\,-\frac{b}{{\small2}a}$$$.

 September 27th, 2011, 09:30 PM #10 Member   Joined: Aug 2009 Posts: 69 Thanks: 0 Re: Difficult parabola equation changed.

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