Fun with coordinate system

Arnav Zek
3 min readApr 25, 2019



foci: Imagine a fat ellipse, every ellipse has two foci, take there distance to a point P on the ellipse. because it’s a foci the addition of these distances will be constant (that’s how we define a foci, we will later talk about how to find them and why are they unique)

how much would it be? let’s take the point P to the end of the major axis (in this case: x) now try to think what would be the resultant sum of the distance.

won’t it be equal to the total stretch on x axis or in other world twice the major radius

simple right

focal length: a² — b² | calculated from center of the ellipse

Latus rectum (a chord through the focus, perpendicular to major axis): 2b²/a

eccentricity (how close is the shape to the circle in shich c/a = 0 ) : c/a

What about hyperbola

Imaging hyperbolas as ellipses whose left part is moved to the right and right part moved to the left

that is what will happen if you start decreasing the value of (X²/a²) or y²/b² from (X²/a² + y²/b² = 1) which is the equation of ellipse and you will find the the equation of hyperbola (X²/a² -y²/b² = 1) or ( -X²/a² + y²/b² = 1)


can you find how far the central vertex will be from the center, use the same method we used in case of ellipse. just in this case it will difference of the distances

but how would you find the foci?? we just did!

we just need to go backwards

focal length: a²+b²

Latus rectum (a chord through the focus, perpendicular to transverse axis)

Conjugate axis: It is a line perpendicular to the transverse axis and passing through the mid-point of two vertices and its length is 2b


pick a point (any point, it will be our focus. so, focus ) and a straight line (in any direction). parabola is the locus of all the point, which is the same distance from the line and that point

isn’t that simple, let’s put that in a formula

remember: distance of the focus from any point on the parabola will be same as the distance of directrix from that point (as we discussed above)

for simplicity sake, we will take our parabola to the center and our directrix parallel to the x axis. so that we can say

directrix and focus are same distance from vertex , taken as p (focal length) naturally ( distance formula)

√(x-0)+(y-p) = y+p

y+p is the total distance of directrix from (x,y) on parabola

after simplification, general equation of parabola: y = x² /4p

focal length (p) = x²/4y

Latus rectum (a chord through the focus, perpendicular to directrix)


section formula

point of intersection: (x1m2 + x2m1/m+n , y1m2 + y2m1 / m+n)

remembering image

x1 x2 y1 y2

\/ \/

m1 m2 m1 m2

just cross multiplying



Arnav Zek

Freelancer . Fulstack Dev . Game Dev