Ellipse: Properties Proved without Calculus

    Theorem 1. Defn. (2) implies defn. (1).
    Proof. Let F, F' be the given foci, at distance 2 c apart, and let FP + PF' = 2 a, where a > c. Chose coordinate axes so that F is (-c, 0) and F' is (0, c). Let A be (-a, 0) and B be (a, 0), then FA + F'A = 2 a, and FB + F'B = 2 a, so that A and B lie on the ellipse. Let e = c/a, and let A' be the point (-a/e, 0), B' be the point(a/e, 0). Let P be any point (x, y) on the ellipse, and let Q be its projection on the x-axis. Let L, L' be the lines ^ x-axis through A', B'. Let the line L" through P parallel to x-axis cut L, L' at N, N'. We shall prove FP / NP = e.
     Notice FP + F'P = 2 a = AQ + BQ = e (A'Q + B'Q) . . . . (a)
Also FP² - F'P² = FQ² - F'Q², so that
FP - FP' = (FQ - F'Q) × (FQ + F'Q) / (FP + F'P)
= (A'Q - B'Q) × 2 c/2 a = e (A'Q - B'Q) . . . . (b) Solving (a) and (b) then gives FP = e NP and F'P = e N'P. Thus F and L serve as focus and directrix defining the ellipse by FP = e NP for all point P, and so do F' and L' also.

    Theorem 2. Defn. (1) implies defn. (2).
    Proof. Choose coordinates so that the focus is F(p,0), and the directrix is the y-axis. Eccentricity is e.
    Lemma. Any line L: y = y₀, will intersect the ellipse K at two points at most, whose midpoint lie on a fixed line parallel to the y-axis.
    Proof: Suppose a point P(x, y₀) is on L Ç K, then let N be the point (0, y₀). Then FP² = e² NP² gives
(x - p)² + y₀² = e² x², or
(1 - e²) x² - 2 p x + (p² + y₀²) = 0 . . . . (i)
This equation has at most 2 roots for x, and the average of the roots is p/(1 - e²), which is a constant independent of y₀. Thus the midpoints of the two intersection points of L and K lie on the line
L': x = p', where p' = p/(1 - e²).
This means the ellipse is symmetric about L' as well as about the x-axis. Lemma is proved.
    Observe if y₀ is too big, then (i) has no roots, because the discriminant in a quadratic formula is
p² - (1 - e²) (p² + y₀²), = e² p² - (1 - e²)y₀².
Then the maximum value of y₀ is e p/Ö(1 - e²) in order for L Ç K to be non-empty.
     Let F' be the mirror image of F across L', and L" the line x = 2 p', so that L" is also the mirror image of y-axis across L'. If L intersects K at two points P(x₁, y₀) and P'(x₂, y₀) where x₁ and x₂ are roots of (i), and L intersects y-axis at N and L Ç L" = N', then due to symmetry across L', F'P = FP' = e NP' = e N'P.
\FP + F'P = e NP + e N'P = 2 e p', which is a constant. This proves that P lies on the curve defined by Defn. 2. QED
    Note: As a special case, if y₀ = 0, then the solutions to (i) are
p/(1 + e) and p/(1 - e).
These points A(p/(1 + e), 0) and A'(p/(1 - e), 0) are mirror images of each other across L'.

    Theorem 3. Defn. 1 implies Defn. 3.
    Proof. Using the same diagram as Thm. 2, Let P be (x, y), then
(x - p)² + y² = FP² = e² NP² = e² x², i.e.,
(1 - e²) x² - 2 p x + (p² + y²) = 0 . . . . (i), i.e.,
(1-e²) [x - p/(1-e²)]² + y² = p² e² /(1-e²), i.e.,
[(1-e²)/p e]² [x - p/(1-e²)]² + [Ö(1-e²)/p e]² y² = 1,
which is of the form in defn. 3, where the center of the ellipse is (p/(1-e²), 0)
and the major and minor radii are a = p e / (1-e²) and b = p e / Ö(1-e²).

    Theorem 4. Defn. 2 implies Defn. 3.
    Proof. [I avoid squareroot sign!] Let P(x,y) be such that
FP + F'P = 2 a . . . . (i) where F(-c,0), F'(c,0) are the foci. Let e = c/a. Let Q be the projection of P on x-axis.
Then from Pythagoras' theorem FQ² - F'Q² = FP² - F'P² .
\ 2 c . 2 x = (FQ + F'Q) (FQ - F'Q) = (FP + F'P) (FP - F'P) = 2 a (FP - F'P).
\ FP - F'P = 2 c x / a = 2 e x . . . . (ii).
Adding (i) and (ii) gives FP = a + e x.
FP² = FQ² + PQ² then gives (a + e x)² = (c + x)² + y², i.e.,
(1-e²) x² + y² = a² (1 - e²), i.e.,
x²/a² + y²/b² = 1 where b² = a² (1 - e²) = a² - c².

    Theorem 5. Defn. 3 implies defn. 1.
    Proof. Let P(x,y) satisfy x²/a² + y²/b² = 1, where a > b. Let c = Ö(a² - b²). Let F be (-c, 0), and the line L be x = a /e where e = c/a.
Then FP² = (c + x)² + y² = (c + x)² + b² (1 - x²/a²)
= (c² + b²) + 2 c x + x² (1 - b²/a²)
= a² + 2 c x + (c x / a)² = (a + c x / a)² = (a + e x)²
= e² (a/e + x)² = e² PN² where N is the projection of P on L.
\ FP = e PN.

    Theorem 6. Defn. 3 implies defn. 2.
    Proof. Let P(x,y) satisfy x²/a² + y²/b² = 1, where a > b. Let c = Ö(a² - b²). Let F be (-c, 0) and F' be (c, 0). Let e = c/a. We need to prove PF + PF' is constant.
     By Pythagoras' thm., PF² = y² + (c + x)². But y² = b² (1 - x²/a²), so that
PF² = b² - x² b²/a² + c² + 2 c x + x²
= x² c² / a² + 2 c x + a² = x² e² + 2 e a x + a² = (a + e x)².
\ PF = a + e x.
     Similarly, PF'² = y² + (c - x)² = . . . = (a - e x)², so that PF' = a - e x.
\ PF + PF' = 2 a, a constant.

    Theorem 7. Let F, F' be the foci of an ellipse K, then for each point P on K, the tangent at P makes equal angles with PF and with PF'.
    Proof.
    Method 1. Assume K is x²/a² + y²/b² = 1 . . . . (i)
and F(-c,0), F'(c, 0) are the foci, where c² = a² - b². Let P be (x, y) and let x = a cos t for some 0 £ t < 2 p, then we pick y = b sin t so that (i) is satisfied. As t moves from 0 to 2 p, P moves one anticlockwise revolution along K. [t is called a parameter for K, so that each value of t represents exactly one point on K].
    Let L be the tangent line at P(a cos t, b sin t), then
its slope = dy/dx = (dy/dt) / (dx/dt) = b cos t / -a sin t. Hence
the slope of the normal L' at P (the line perpendicular to the tangent line at P) = -1 / above = (a/b) tan t. Hence by the "one point + slope" form of eqn. of str. line, the eqn. of L' is
    y - b sin t = (a/b) tan t (x - a cos t). Let it cut the x-axis at Q where y = 0,
\ - b sin t = (a/b) tan t (x - a cos t),
\ x = a cos t - (b²/a) cos t = e c cos t where e = c/a.
\ area DPFQ / area DPF'Q = QF / QF' = ( c + e c cos t) / (c - e c cos t)
= (a + e a cos t)/(a - e a cos t) = (a + e x)/(a - e x) = PF / PF'.
\ Q is equidistant from lines PF and PF'. \ PQ bisects ÐFPF'.
Since PQ ^ L, \ L makes equal angles with PF and PF'.

    Method 2 (no calculus, no coordinate geometry !!!) Let the tangent line at P be TPT', where ÐFPT and ÐF'PT' both < p/2. Suppose ÐFPT ¹ ÐF'PT'. Let F" be the mirror image of F' across the line TPT'. Then FPF" is not a str. line, hence the line FF" crosses TPT' at some point P" different from P. P" is outside the ellipse because TPT' is a tangent. But then P"F + P"F' = P"F + P"F" = FF" < FP + PF" = FP + PF', which is possible only if P" is inside the ellipse. This is absurd. Therefore ÐFPT can only = Ð F'PT'.

    Theorem 8. If P(x', y') is a point on the ellipse K: x²/a² + y²/b² = 1 . . . (i)
then the tangent at P is x x'/a² + y y'/b² = 1 . . . (ii)
    1st Proof. (ii) represent some line L which cuts K at points (x, y) satisfying (i) and (ii) simultaneously. Eliminating y, from (ii) we get y = b²/y'(1 - x x'/a²). Substitute into (i), we get
x²/a² + (1 - x x'/a²)² b²/y'² = 1, i.e.,
y'²x² a² + b² (a⁴-2 a²x x' + x² x'²) = a⁴ y'², i.e.,
x² (y'²a² + b² x'²) - 2 a² b²x'x + a⁴(b² - y'²) = 0, i.e.,
b²x² - 2 b² x'x + a²(b²-y'²) = 0 . . . . (iii)
The discriminant of this quadratic equation in x is b⁴ x² - b² a²(b² - y'²) = b² (b² x'² + a² y'² - a² b²) = 0.
Hence (iii) has a repeated root, which means the line (ii) cut (i) at only one point, hence (ii) represents the tangent line.
    2nd Proof.. Consider a new variable v = a y / b. Define a mapping f mapping every point (x, y) in the plane to the point (x, v) where v = a y / b, which stretches the plane like an elastic membrane in the y-direction. Under this mapping, the ellipse x²/a² + y²/b²= 1 becomes x²/a² + v²/a² = 1, which is a circle.
The point P(x', y') becomes the point P'(x', a y'/b) where
the tangent line L' is obviously v - a y'/b = (-x'/(a y'/b))(x - x'), i.e.,
v - a y'/b = (b x' / a y') (x' - x). Inverse mapping of this line under f^-1 then gives
a y /b - a y'/b = (b x'/a y') (x' - x), i.e.,
a² y y' + b² x x' = b² x'² + a² y'² = a² b², i.e.,
x x'/a² + y y'/b² = 1.

    Theorem 9 (Existence of a Pole for a circle). Let L be a line not intersecting a circle Z of radius r and center O. Let OMN be a line perpendicular to L, intersecting L at N, and that OM × ON = r². Then for all points P on L, the line (called the polar of P) joining the points of contact of the two tangent lines from P to the circle, will always go through M.
    Proof. Let OP cut QR at S, and let ON cut QR at A. Notice ASPN and ROPN are cyclic quadrilaterals because of right angles. Hence ÐORS = ÐOPR = ÐONR. \DOAR and DORN are similar, so that OA : OR = OR : ON, so that OA × ON = r². Hence A = M. \Regardless of P's position on L, QR cut ON at the point M, which is called the pole of L relative to Z.

    Theorem 10 (Existence of a Polar for any point in an ellipse). Let Z be an ellipse and M any point inside. Let a variable chord QR constantly go through M. Then the locus of the intersection point of the two tangents at Q and R is a straight line L, called the polar of M. In particular, if M is a focus, then L is the directrix.
    Proof. [We purposely avoid a messy proof with coordinate geometry.] Let the ellipse be x²/a² + y²/b² = 1. let a mapping f take every point (x, y) in the plane to the point (x, v) where v = a y / b. This is merely a dilatation in the y-direction. Then under f straight lines are mapped to straight lines, the ellipse to a circle, and tangents mapped to tangents. Then the problem becomes that of a circle, where, as in the previous theorem, QR is a variable chord going through the fixed point M on a fixed line ONM where OM×ON = r². Then the tangents at Q,R intersect at some point P on the line L through N perpendicular to OMN. Mapping back under the inversion yields the truth of this theorem. In particular, if M is the focus (a e, 0), the mapped circle has radius a, and the directrix is x = -a/e, intersecting x-axis at the point N(-a/e, 0), then ON×OM = a², proving the fact that the directrix is the polar of the focus. QED

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