# Find the points where lines cross parabolas or hyperbolas

Normal to a parabola at point P, (ap2, 2ap): where it crosses the curve again

On the parabola y = 2at and x = at2

Therefore dy/dt = 2a; dx/dt = 2at; dy/dx = 1/t;
and slope of normal = −t

Sometimes in Edexcel mark schemes, they write this as a matter of dy/dp and dx/dp. Strictly speaking this is wrong. In this context p is a constant. It describes a single point. You can’t find derivatives by looking at the curve at a single point. You have to look at a neighbourhood around that point. Here, weirdly, the way to find the slope of the normal at (ap2, 2ap) is to find a formula which tells us the slope of the normal everywhere.

At P, t = p, so slope of normal = −p

So equation of normal is

(y − 2ap) = − p (x − ap2) ………… [1]

This line crosses the curve at points with timestamp t where

(2at − 2ap) = − p (at2 − ap2)

We could instead have solved a set of simultaneous equations, y2 = 4ax and equation [1]. But it’s neater to take our information about the curve in the form “the typical point is (at2, 2at)”, which includes exactly the same information as “the equation of the curve is y2 = 4ax”, and find which point by plugging y=2at and x=at2 into the equation [1]. We now have one equation in one variable, t, instead of two equations in two variables. Half the trouble.

2(t − p) = − p (t2 − p2)

Minimise the problem which Tegan pointed out − “lots of letters in the equation” − by cancelling out a

pt2 + 2t + … = 0

It’s an equation in t, where p functions only as a constant. Write it in the usual way we write quadratic equations, with t2 term first, t term second, constant term third. The constant term actually doesn’t matter for the rest of this calculation, which is why I’ve written just … there.

t2 + (2/p)t + … = 0

Divide through by p so as to write the quadratic in standard form (coefficient of t2 set to equal one)

Sum of roots = −2/p

One of the roots is p, so the other root is −(2/p) − p

We know one of the roots is p, because the equation is for timestamps of points where the line crosses the curve. The line is the normal at P, so one of the roots MUST BE the timestamp at P, which is p. Just as if you leave your house to go out one evening, one of the points on your route MUST BE home.

To find the coordinates of the other point where the normal crosses the curve, just plug t = −(2/p) − p into the formula for the typical point, (at2, 2at).

Normal to a hyperbola at point P, (c/p, cp): where it crosses the curve again

On the hyperbola y = c/t and x = ct

Therefore dy/dt = −ct−2; dx/dt = c; dy/dx = −t−2;
and slope of normal = t2

At P, t = p, so slope of normal = p2

So equation of normal is

(y − c/p) = p2 (x − cp)

This line crosses the curve at points with timestamp t where

(c/t − c/p) = p2 (ct − cp)

(1/t − 1/p) = p2 (t − p)

t2 − (p − p−3)t + … = 0

(Multiplying through by t, dividing through by p2, and rearranging)

Sum of roots = p − p−3

One of the roots is p, so the other root is − p−3

Line from a point P on a parabola (ap2, 2ap) through the focus (a,0): where it crosses the curve again

Equation of line is

(y − 0) = 2ap(a − ap2) (x − a)

or

y = 2p(1 − p2) (x − a)

This line crosses the curve at points with timestamp t where

2at = 2p(1 − p2) (at2 − a)

or: t2 − (p − 1/p) t + … = 0

Dividing through by 2ap, multiplying through by (1 − p2), and rearranging

Sum of roots = p − 1/p

One of the roots is p, so the other root is −1/p