# Geometrical proof for length of chord passing through vertex of parabola

In a parabola $$y^2=4ax$$ , the length of focal chord making an angle $$\theta$$ with the x - axis is $$4acosec^2\theta$$ . If a chord is drawn parallel to that focal chord which passes through vertex of parabola at (0,0) , it's length comes out to be $$4acosec^2\theta cos\theta$$ , it's quite easy to prove this using parametric coordinates for the parabola , I'm looking for an intuitive geometric demonstration that AB=A′B′.The equality certainly holds but I feel there must a visual way to show it.I have attached two geometrical 'viewpoints' from which i am seeing the scenario.

Changing the notation a bit, let $$\overline{AB}$$ be a chord through the parabola's focus, $$F$$, and let $$\overline{UV}$$ be a parallel chord through the vertex, $$V$$. Let $$\overleftrightarrow{A^\prime B^\prime}$$ be the parabola's directrix, and let $$C$$ be the fourth vertex of rectangle $$\square AA^\prime B^\prime C$$.

Writing $$a := |AA^\prime| = |AF|$$ and $$b := |BB^\prime| = |BF|$$ (and, without loss of generality, assuming $$a\geq b$$), we see that $$|BC| = a-b$$.

Now, recall that the midpoints of parallel chords of a parabola lie on a line parallel to the axis of that parabola. Consequently, if $$M$$ and $$N$$ are the midpoints of $$\overline{AB}$$ and $$\overline{UV}$$, respectively, then defining $$d := |UN|=|VN|$$, we have $$b + d = |BM| = \frac12|AB| = \frac12(a+b) \quad\to\quad d = \frac12(a-b) \quad\to\quad |UV| = |BC|$$

• Brilliant , just the kind I was looking for. Feb 3 '19 at 10:46

This isn't the approach you're taking, but it's pretty straightforward.

In the figure, $$V$$, $$F$$, and $$\overleftrightarrow{DQ}$$ are the the parabola's vertex, focus, and directrix. Recall that, by the defining property of the parabola, $$|PF| = |PQ|$$.

In $$\triangle RPF$$, $$|PR|^2 = \color{red}{|PF|^2} - |RF|^2 = \color{red}{|PQ|^2} - |RF|^2 = (|PQ|+|RF|)(|PQ|-|RF|)$$ $$c^2\sin^2\theta = 2 c\cos\theta \cdot 2a$$ $$c = 4a\cos\theta \csc^2\theta$$

• As I said , we can prove this with alternate ways but , I am not sure if I am putting this correctly , I was looking for a "cut and paste" kind of connection between AB and A'B' as marked in my diagram..the sort of visual proof we have Pythagoras theorem , I strongly feel there should be something. Feb 3 '19 at 7:58
• @ADITYAPRAKASH: Well, I acknowledged that this wasn't quite what you wanted; perhaps it might be helpful to others to see the length formula explicitly verified. That said, I believe I was a little confused by your question; I had interpreted it as looking to verify $AB=A'B'$ en route to showing the specific formula for the length of the vertex chord. I realize now that "all you want" is a geometric demonstration of the equality; the length formulas themselves are irrelevant, except insofar as they assure us that the equality is true. Is this correct?
– Blue
Feb 3 '19 at 8:28
• Exactly , perfectly put. Feb 3 '19 at 8:30
• Then might I suggest rephrasing your question a bit? Display the image and write something like "I'm looking for an intuitive geometric demonstration that $AB=A'B'$. I happen to know (from analyzing a parameterization of the parabola) that the focal chord length is $4a\csc^2\theta$ and that $AB=4a\csc^2\theta\cos\theta$. So, the equality certainly holds, but I feel there should be a nice visual way to show it."
– Blue
Feb 3 '19 at 8:48