Some Examples on Parabola.

    Problem 2. Prove that a chord of a parabola, which goes through its focus, has the tangents at its endpoints intersecting each other perpendicularly on the directrix.
    Proof. Let the parabola be K: y² = 4 a x, and PFQ be a chord going through its focus F(a, 0). Let P', Q' be the projections of P, Q on the directrix x = -a. [We can prove the tangents at P, Q are perpendicular to each other easily using (1): Let the tangents at P, Q meet at T. Then TP bisects ÐQPP' and TQ bisects ÐPQQ', due to problem (1). Then ÐTPQ + ÐTQP = (½)ÐQPP' + (½)ÐPQQ' = p/2. \ÐPTQ = p/2.] (But in an exam, one cannot quote results proved in another problem.)
    Let P be (x₁, y₁) and Q be (x₂, y₂). Then F lies on PQ if slope of PF = slope of PQ, or
(y₁-0) / (x₁-a) = (y₁-y₂) / (x₁-x₂), i.e.,
y₁ (x₁-x₂) = (y₁-y₂) (x₁-a), i.e.,
y₁ (y₁²-y₂²) / 4 a = (y₁-y₂) (y₁²-4 a²) / 4 a, i.e.,
y₁ (y₁ + y₂) = y₁²-4 a², i.e.,
y₁y₂ = -4 a² . . . . (4).
    Now the slope of K at P is dy/dx = (dx/dy)^-1 = (2 y₁ / 4 a)^-1 = 2 a/y₁, hence the tangent line at P is
y-y₁ = (2 a/y₁) (x-x₁), or y = 2 a x/y₁ + y₁ / 2 . . . . (2)
The slope of this tangent at P is 2 a / y₁. Similarly, the slope of the tangent at Q is 2 a / y₂, so that the product of slopes is 4 a²/y₁y₂ = -1 due to (4). Therefore the tangents are perpendicular to each other.
    Again, their equations are
y = 2 a x/y₁ + y₁ / 2 and y = 2 a x/y₂ + y₂/2.
They intersect each other at some point T(x, y) satisfying these equations.
Subtracting, we get 2 a x (y₂-y₁) / y₁y₂ + (y₁-y₂)/2 = 0, i.e.,
4 a x / y₁y₂ = 1. Together with (4), we get
4 a x /(-4 a²) = 1, i.e., x = -a.
This means T lies on the directrix. QED

    Problem 3. Given a point A(0,2), B(0,4) and a line L: y = x + 1. Find a parabola K with L as axis, and going through A and B.
    Solution. First, let us move origin to some point on L, such as C(0,1), with y=y'+1. Then in (x,y') coords, A is (0,1), B is (0,3) L is y' = x.
Next we rotate axes 45 degrees with x = (x" - y")/Ö2 y' = (Y" + x")/Ö2,
or x"=(x + y')/Ö2, y"=(y'-x)/Ö2. Then in (x", y") coords, A is (1/Ö2, 1/Ö2), B is(3/Ö2, 3/Ö2).
     A parabola in (x",y") system with x"-axis as axis of symmetry is y"² = 4 a (x"-h). We want it to pass through A and B, from which we get 2 equations to solve for a and h.
From A we get 1/2 = 4 a (1/Ö2-h) . . . . (i)
From B we get 9/2 = 4 a (3/Ö2-h) . . . . (ii).
(ii)-(i) then gives 4 = 4 a Ö2, so that a = 1/Ö2. \ h = 3Ö2/8, and 4 a h = 3/2.
\The equation of the parabola is y"² = 2Ö2×x" - 3/2 ,
i.e., (y'-x)²/2 = 2(x+y')-3/2
i.e., (y-1-x)² = 4(x + y-1)-3,
i.e., (y-x)² -2(y-x)+1= 4 x + 4 y - 7,
i.e., x²-2 x y + y² -6 y -2 x + 8 = 0.QED

    Problem 4 (pole of a chord). Given a parabola K: y² = 4 a x . . . . (1) Let R be the midpoint of any chord PQ. Let the tangent lines at P and Q meet at S, called the pole of the chord. Prove that (a) SR is parallel to the x-axis, (b) the midpoint T of RS lies on K, and (c) the tangent at T is parallel to RS.
    Proof. Let P, R, Q have y-coords k+h, k, k-h respectively. Then P is ((k+h)²/4 a, k+h), Q is ((k-h)²/4a, k-h). The x-coord of R is the average of those of P, Q. Hence R is (k²+h²/4 a, k). The tangent at any point W(w²/4 a, w) is y - w = (2 a / w) (x - w²/4 a) = 2 a x / w - w/2, i.e.,
y = 2 a x / w + w/2. Hence the tangents at P, Q are
y = 2 a x / (k+h) + (k+h)/2 . . . . (3) and y = 2 a x / (k-h) + (k-h)/2 . . . . (4) which intersect at S.
(3)-(4) then gives the x-coord of S as 2 a x (-2 h) / (k²-h²) + h = 0, or x = (k+h)(k-h) / 4 a.
Substituting in (3) or (4) then gives y = k, proving SR is parallel to x-axis.
     Averaging the x-coords of R and S then gives T as (k²/4 a, k), which satisfies the equation (1), proving T lies on K.
     Finally, the tangent at T is y = 2 a x / k + k / 2, which has slope 2 a / k.
The slope of PQ is [(k+h)-(k-h)]/ [(k+h)²/4 a - (k-h)²/4 a] = 2 h / (4 k h / 4 a) = 2 a / k.
Hence the tangent at T is parallel to PQ. QED

    Defn. Given a parabola K, the set of points in the plane each lying on some chord of K, is called the interior of K.

    Problem 5 (Polar of a point). Let R be any point in the interior of a parabola
K: y² = 4 a x. . . . . (1) For each chord PQ going through R, let the tangents at P and Q intersect at S. Then for all such chords, the points S lie on some straight line, parallel to the chord with R as midpoint, called the polar of R relative to K.
    Proof. Any interior point R of K can be represented as ((k²+h²)/4 a, k), because the line through R parallel to the x-axis hits K at (k²/4 a, k) whose x-coord < x-coord of R.
    Any chord through R is y-k = m (x- (k²+h²)/4 a) . . . . (2), for some m.
Hence (1) and (2) intersect at (x, y) where y-k = m (y²/4 a - (k²+h²)/4 a) , i.e.,
y² - 4 a y / m + 4 a k / m - k² - h² = 0. This has two roots y₁ and y₂ whose sum is 4 a/m and product is 4 a k/m-k²-h². So they represent two endpoints of the chord PRQ, where
P is (y₁²/4 a, y₁) and Q is (y₂²/4 a, y₂).
    The two tangent lines at P,Q are y = (2 a / y_i) x + y_i/2, for i = 1, 2. . . . . . (3),(4).
They intersect at S, where the x-coord is obtained by (3)-(4), giving
2 a x (y₂-y₁)/y₁y₂ = (y₂-y₁)/2, or x = y₁y₂/4 a. From (3), we get y = (y₁+y₂)/2.
Thus S(x,y) = (y₁y₂/4 a, (y₁+y₂)/2) = (k/m-(k²+h²)/4 a, 2 a/m)
\S lies on the curve x = k/m-(k²+h²)/4 a, y =2 a/m. Eliminate m,
\ 4 a x = 2 k y - (k²+h²), which is a straight line with slope 2 a /k. In particular, if y = k, then x = (k+h_(k-h)/4 a, same as the point S of the previous problem, where the chord with R as midpoint has endpoints ((k+h)²/4 a, k + h) and ((k-h)²/4 a, k-h), with slope 2 a /k also. QED

    Problem 6 (mid points of parallel chords). Given any non-zero number m, prove that the chords of K: y² = 4 a/x with slope m have all their midpoints lying on some line L that cuts K at some point T whose tangent also has slope m. For each chord, the tangents at the endpoints also intersect at some point on L.
    Proof. Let such a chord be y = m x + c. It cuts K at (x, y) where
y² = 4 a (y-c)/m, whose roots y₁ and y₂ satisfy
y₁+y₂ = 4 a/m and y₁y₂ = 4 a c/m. These two roots represent the endpoints of the chord,
P(y₁²/4 a, y₁), and Q(y₂²/4 a, y₂). \Their midpoint R(x, y) lies on
y = (y₁+y₂)/2, or y = 2 a/m. Thus the midpoints of all the parallel chords lie on the line L: y = 2 a/m.
    The tangent line at P is y = 2 a x/y₁ + y₁/2 . . . . (3) and that at Q is y = 2 a x/y₂ + y₂/2. Subtracting, we get the x coord of their intersection point S as
2 a x(y₂-y₁)/y₁y₂ = (y₂-y₁)/2, or x = y₁y₂/4 a = (4 a c/m) /4 a = c/m.
Substituting x = y₁y₂/4 a into (3) gives the y-coord of S = (y₁+y₂)/2 = 2 a/m, so that S lies on L.
     Finally, L cuts K at ((2 a / m)²/4 a, 2 a / m) or (a / m², 2 a/m), whose tangent has slope
2 a / (2 a / m) = m, same as the slope of the parallel chords. QED

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