The parent proof throws a lot of notation at you in one breath: y2=4ax, F=(a,0), the directrix x=−a, the slope y12a, angles α and β, and the formula tanθ=1+m1m2m1−m2. We will earn each one, in an order where every new symbol only leans on symbols already built.
Look at the figure below. The orange dot is at (3,2): walk 3 steps right, then 2 steps up. That single label (3,2)is a picture — an address on the grid.
Why the topic needs this: the whole proof lives on this grid. Points like P=(x1,y1) and F=(a,0) are just addresses, and "the ray goes from P to F" means "draw the line joining two addresses."
Why the topic needs this: the incoming star-light is modelled as many horizontal rays. "Horizontal" is the input condition of the whole reflective property.
Before the parabola, we need one careful idea: what does "how far is a point from a line" even mean?
The figure shows why: the perpendicular drop (magenta) is shorter than any slanted path (dashed). For a vertical directrix line x=−a, dropping a perpendicular is easy — it is just a horizontal move, so M has the same y as P and sits on the line at x=−a.
Why the topic needs this: the incoming ray runs horizontally straight at M. The whole reflective property is glued to the distance PM, so "distance to a line" must mean the perpendicular one — nothing else.
Now meet the shape by its rule, then earn the equation.
The figure shows this equal-distance balance: the two coloured segments PF and PM always have the same length, for every point P.
Let us turn the words into algebra. Choose the vertex at the origin, focus F=(a,0) (a distance a right), and directrix the vertical line x=−a (a distance a left). Take a general point P=(x,y).
Step A — write PF. The distance between two points (x,y) and (a,0) comes from the Pythagoras rule (a right-triangle with legs x−a and y):
PF=(x−a)2+(y−0)2.
Step B — write PM. From §4, the foot on the vertical line x=−a is M=(−a,y) (same height as P). So PM is just the horizontal gap:
PM=∣x−(−a)∣=∣x+a∣=x+a(since x≥0 on the curve).
Step C — impose PF=PM and square (squaring removes the square-root and the absolute value, keeping both sides positive):
(x−a)2+y2=(x+a)2.
Step D — expand and cancel. Using (x±a)2=x2±2ax+a2:
(x−a)2x2−2ax+a2+y2=(x+a)2x2+2ax+a2.
Cancel the x2 and a2 that appear on both sides:
−2ax+y2=2ax⟹y2=4ax.
Recall Why is it
y2 and not y?
Because for one x there are two points — one up, one down — symmetric across the axis. Squaring y lets both +y and −y satisfy the same equation. A single x gives two y-values ::: because y and -y both square to the same y^2
The figure places all three. Notice the focus and directrix are mirror-balanced about the vertex — one step right, one step left. That balance is exactly what let the x2 and a2 cancel in §5.
Why the topic needs this: the focus F is the "meeting door" all rays run to. Every worked example first hunts down a.
A straight line has one slope everywhere. A curve bends, so its steepness changes point to point.
Deriving the tangent slope. We want the slope of y2=4ax. The trick is implicit differentiation: differentiate both sides with respect to x, remembering that y is secretly a function of x, so a "chain-rule tag" dxdy tags along whenever we differentiate a y.
Differentiate the right side 4ax with respect to x: the rate of change of 4ax is 4a (each step right adds 4a).
Differentiate the left side y2: the rate of change of "something squared" is 2×(that something)×(its own rate), i.e. 2y⋅dxdy.
Setting the two rates equal:
2ydxdy=4a⟹dxdy=2y4a=y2a.
At the chosen point P=(x1,y1) this is the tangent slopemt=y12a.
Why both appear: physics says "angle in = angle out about the normal." The parent proof instead measures equal angles about the tangent — the two statements are the same because tangent and normal are perpendicular, and the tangent slope y2a is simply easier to compute.
