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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

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    $\begingroup$ 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. $\endgroup$
    – Blue
    Commented Nov 13, 2020 at 11:30
  • $\begingroup$ Oh nice, is my proof correct by the way? $\endgroup$
    – Babu
    Commented Nov 14, 2020 at 5:59

1 Answer 1

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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 $VF=VZ=Lr/4.$

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