# Relation of the length of latus rectum with the distance between focus and vertex and distance to focus from directix

In this website, the following statement is written

Length of latus rectum is twice the distance between focus and vertex or four times the distance of focus from the directrix.

I thought of proving the statement for a parabola whose focus lies on the $$y$$ axis and this would prove the geometric idea for all other parabolas since all other are just shifted/stretched variations of the one in consideration. The proof is presented in the following section.

Set of points equidistance from line and point. For a parabole opening up like a U in the x-y plane, say it is parallel its line is $$y=c$$ and the point is $$(a,b)$$ then the distance from these two are the same.

$$(y-c)^2 = (x-a)^2 + (y-b)^2 \tag{1}$$

Now, the for the points which are same vertical height as directrix, plug $$y=b$$,

$$a \pm (b-c) = x$$

Hence the distance between the two points:

$$l= 2 (b-c)$$

"The length of latus rectum is twice the perpendicular distance between directrix and focus"

But the statement says that it is four times, so is it wrong or did I make mistake?

For the second part, we can find the maxima of $$y$$ is when $$x=a$$ by differentiation and hence,

$$y-c = -(y-b)$$

$$2y = b-c$$

Hence, the vertex is given as:

$$y = \frac{b-c}{2}$$

Now since focus at (a,b) the distance between vertex and focus :

$$d = \frac{b+c}{2}$$

Which doesn't even seem to be related to original equation

• The statement asserts that $$2\times(\text{focus-to-vertex}) = 4\times(\text{focus-to-directrix})$$ which cannot be true, as the vertex lies between the focus and directrix. So, the author inadvertently swapped "directrix" and "vertex" (or "twice" and "four times"); these things happen. Note that in the working of the problem, the author correctly calculated the latus rectum by merely doubling the focus-to-directrix distance.
– Blue
Commented Nov 13, 2020 at 11:30
• Oh nice, is my proof correct by the way?
– Babu
Commented Nov 14, 2020 at 5:59

For the parabola $$y^2=4ax$$, the equation of Latus rectum is $$x=a$$ and end points of Lr are $$(a,2a)$$ and $$(a,-2a)$$ so $$Lr=4a$$. The vertex is $$V(0,0)$$ directrix is $$x=-a$$ The point of intersection of directrix and x-axis is $$Z(-a,0)$$ so $$VZ=a$$. Focus is $$F(a,0)$$ so $$VF=a$$. So finally $$VF=VZ=Lr/4.$$