Parabola

GIKPKC7 94107

Parabola I

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Introduction 10/8/98

Terms: Normal Cartesian Parametric

Perpendicular to Equations involving x & y E.g. x2 = 4.a.y Equations involving x & y but written in term of a third variable

A parabola has a minimum value if a > 0

A parabola has a maximum value if a < 0

Quadratic Inequalities: For any curve in the number plane:  If, a.x2 + b.x + c = 0 on the x-axis 2  If, a.x + b.x + c > 0 above the x-axis  If, a.x2 + b.x + c < 0 below the x-axis

Roots of Quadratic Equations:    

Roots = x-intercepts of a graph 2 roots (negative and positive regions) of a parabola = Indefinite Function 1 roots of a parabola = Not Called Anything No roots of a parabola and: a > 0 it’s known as a Positive Definite a < 0 it’s known as a Negative Definite

Luke Cole

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GIKPKC7 94107

Parabola I

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The Discriminant:

 Under the square root sign of the quadratic formula = discriminant =  Equation:  = b2 – 4.a.c       

If,  < 0 no real roots If,  = 0 one real root (two equal roots) If,  > 0 two real roots (two unequal roots) If,  = perfect square = rational If,  = not a perfect square = irrational For, a.x2 + b.x + c > 0 <0 a>0 2 For, a.x + b.x + c < 0 <0 a<0

Sum & Product of Roots: General form: x2  ( + )x + . = 0 Proof: If  and  are two roots of a quadratic equation then: (x  )(x  ) = 0  x2  ( + )x + . = 0 …[1] Equation:

Equation:

b a Sum of roots  

c a Product of roots  . 

Proof: a.x2 + b.x + c = 0 b c x 2  .x   0 …(2) a a Compare (2) to [1]: x2  ( + )x + . = 0 b c x 2  .x   0 a a b        & a b    a

 . 

c a

Luke Cole

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GIKPKC7 94107

Parabola I

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Quadratic Identities: a1.x2 + b1.x + c1 = 0  a2.x2 + b2.x + c2 = 0 a1 = a2 b1 = b2 c1 = c2

Equation:

Equations Reducible to Quadratics: E.g.

Solve, x 

2 3 x

x2 + 2 = 3.x (x – 2)(x – 1) = 0 x=2 & x=1

A

E.g.

Solve, 32.x – 4.3x + 3 = 0

A

Let, 

u = 3x u2 – 4.u + 3 = 0 (u – 3)(u  1) = 0 u=3 & u=1 Sub, u = 3x into u = 3 & u = 1 3x = 3 & 3x = 1 x=1 & x=0

Tangent’s and Normals:

 To find the gradient of a tangent use calculus (See: The Tangent & Derivative)  To find the normal use the formula: m1.m2 =  1 E.g. Find the equation of the tangent and normal of y = x2 at (3, 9) y = x2 dy  2.x dx Sub x = 3 into 2.x to find tangent m1 = 6  Y – y = m1(X – x) y – 9 = 6(x – 3)  6.x – y – 9 = 0 Finding the Normal  m1.m2 =  1 m2 =  1/6  Y – y = m2(X – x) y – 9 =  1/6(x – 3)  x + 6.y – 57 = 0 Luke Cole

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GIKPKC7 94107

Parabola I

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Parabola as a Locus 

19/8/98 The locus of a point that is equidistant from a fixed point and a fixed line is always a parabola.

Definitions: x2 = 4.a.y Tangent Focus

Chord (0, a)

Focal Length

Focal Chord a (0, 0) Vertex a y=a

Directrix



When the Focal Chord is parallel to the Directrix it is called Latus rectum

Concave Up: 

The locus of point P(x, y) moving such that it is equidistant from (0, a) and the line y =  a is a parabola General Form: (x – h)2 = 4.a(y – k) (h, k) = Vertex a = Focal length Proof: A[h, (k + a)] P(x, y) Since, PA = PB: (x – h)2 + [y – (k + a)]2 = (x – x)2 + [y  (k  a)]2 2 (x – h) = 4.a.y – 4.a.k  (x – h)2 = 4.a(y – k) B[x, (k  a)] y=ka

Luke Cole

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GIKPKC7 94107

Parabola I

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The Alternative Case: 

The locus of point P(x, y) moving such that it is equidistant from (a, 0) and the line x =  a is a parabola General Form: (y – k)2 = 4.a(x – h) Proof: Since, PA = PB: [x – (h + a)]2 + (y – k)2 = [x – (h – a)]2 + (y  y)2 (y – k)2 = 4.a.x – 4.a.h  (y – k)2 = 4.a(x – h) x=ha A[(h + a), k]

B[(h  a), y)

P(x, y)

Parametric Equations of the Parabola: Parametric of x2 = 4.a.y:

x = 2.a.t & t = Parametric

y = a.t2

Proof: x = 2.a.t y = a.t2 Sub (1) into (2):

 x  y  a   2.a  x2 = 4.a.y E.g.

…(1) …(2) 2

Find the cartesian equation of x = 3.t + 1 & y = 2.t – 3 x = 3.t + 1 …(1) y = 2.t – 3 …(2) Sub (1) into (2) 2 .x  2 y 3 3 2.x – 3.y – 11 = 0

Luke Cole

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GIKPKC7 94107

Parabola I

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Chords in Parametric Form: If P(2.a.p, a.p2) and Q(2.a.q, a.q2) are any Two Points on the Parabola x2 = 4.a.y Then Gradient:

m PQ 

Proof: Using gradient formula: a. p 2  a.q 2 m PQ  2.a. p  2.a.q a p  q  p  q  = 2.a  p  q  pq m PQ   2

p q 2

The Equation of the Chord PQ Will Be

General Form: y  ½(p + q)x + a.p.q = 0 Proof: Using the point, gradient formula:  p q y  a. p 2   a  2.a. p   2   y  ½(p + q)x + a.p.q = 0 …[1]

If PQ is a Focal Chord Then

General Form: p.q =  1 Proof: x2 = 4.a.y has a focus (0, a) & the equation of PQ is y  ½(p + q)x + a.p.q = 0 then sub (0, a) into the equation PQ: a.p.q =  a  p.q =  1

Luke Cole

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GIKPKC7 94107

Parabola I

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Tangents in Parametric Form: The Tangent to the Parabola x2 = 4.a.y at P(2.a.p, a.p2) has Equation: y – p.x + a.p2 = 0 Gradient: m=p Proof: Differentiate x2 = 4.a.y to find gradient: dy x  dx 2.a 2.a. p = 2.a  m=p …[2] Use the point, gradient formula to find equation: y – a.p2 = p(x – 2.a.p)  y – p.x + a.p2 = 0 …[3]

The Tangents to the Parabola x2 = 4.a.y at the Points P(2.a.p, a.p2) and Q(2.a.q, a.q2) Intersect at Point: [a(p + q), a.p.q] Proof: Equation of the tangent a P is [3]:  y – p.x + a.p2 = 0 …(1) Equation of the tangent a Q is the same as [3], but has ‘p’s as ‘q’s:  y – q.x + a.q2 = 0 …(2) Now, (1) – (2): P  p.x + q.x + a.p2  a.q2 = 0 Q (q – p)x + a(p  q)(p + q) = 0  x = a(p + q) Sub x = a(p + q) into (1): Z y – p[a(p + q)] + a.p2 = 0  y = a.p.q  Z[a(p + q), apq] …[4]

Luke Cole

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GIKPKC7 94107

Parabola I

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Normals in the Parametric Form: The Normal to the Parabola x2 = 4.a.y at P(2.a.p, a.p2) has: Equation: Gradient:

x + p.y = a.p3 + 2.a.p 1 m p

Proof: Use [2] and the perpendicular line formula: m1.m =  1 p.m =  1 1 m  p Use the point, gradient formula to find the equation:   1 y  a. p 2    x  2.a. p   p   x + p.y = a.p3 + 2.a.p …[5]

The Normal’s to the Parabola x2 = 4.a.y at P(2.a.p, a.p2) and Q(2.a.q, a.q2) Intersect at Point: [ a.p.q(p + q), a(p2 + p.q + q2 + 2)] Proof: Equation of the tangent a P is [5]:  x + p.y = a.p3 + 2.a.p …(1) Equation of the tangent a Q is the same as [5], but has ‘p’s as ‘q’s:  x + q.y = a.q3 + 2.a.q …(2) Z Now (1) – (2): p.y – q.y – 2.a.p + 2.a.q – a.p3 + a.q3 = 0 P  y = a(p2 + p.q + q2 + 2) Sub y = a(p2 + pq + q2 + 2) into (1): x + p[a(p2 + p.q + q2 + 2)] = a.p3 + 2.a.p  x =  a.p.q(p + q)  Z[ a.p.q(p + q), a(p2 + p.q + q2 + 2)]

Luke Cole

Q

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Parabola I

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Tangents in Cartesian Form: The Equation of the Tangent to the Parabola x2 = 4.a.y at P(x1, y1) is General Form: x.x1 = 2.a(y + y1) Proof: Differentiate x2 = 4ay to find gradient: dy x  dx 2.a x m 1  …[6] 2.a Use the point, gradient formula to find the equation: x  y  y 1   1  x  x1   2.a  2.a(y – y1) = x1.x – x12 Since x2 = 4.a.y then, x12 = 4.a.y1: 2.a(y – y1) = x1.x – 4.a.y1  x.x1 = 2.a(y + y1)

Normal’s in Cartesian Form: The Equation of the Normal at P(x1, y1) on the Parabola x2 = 4.a.y is General Form:

  2 .a  y  y1    x  x 1   x1 

Proof: Use [6] and the perpendicular formula:  m1.m =  1  x1    .m  1  2.a   2.a m  x1 Use the point, gradient formula to find the equation:   2.a  y  y1    x  x1    x1 

Luke Cole

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GIKPKC7 94107

Parabola I

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The Equation of the Chord of Contact XY of the Tangents Drawn From the Point P(x1, y1) to the Parabola x2 = 4.a.y is: General Form: x.x1 = 2.a(y + y1) Proof: The points in parametric form would be X(2.a.p, a.p2) and Y(2.a.q, a.q2); so the equation of the chord is [1]:  y  ½(p + q)x + a.p.q = 0 X The point were the tangents intersect is [4]: Y  P[a(p + q), a.p.q] Now comparing the points P[a(p + q), a.p.q] & P(x1, y1): x1 = a(p + q) x pq  1  …(1) a y1 = a.p.q …(2) P Now sub (1) & (2) into [1]:  x1  y  ½.   .x + y1 = 0 a  x.x1 = 2.a(y + y1)

Luke Cole

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Parabola I

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Properties of a Parabola: x = 2at, y = at2 The Tangent to a Parabola at a Given Point is Equally Inclined to the Axis and the Focal Chord Through the Point: FPQ = PQF

i.e.

F

Q

P

F = Focal chord = (0, a) P = Point = (2.a.p, a.p2) Q = Tangent x-intercept FQP =  QPF = 

Z

Proof: Tangent at P is [3]:  y = p.x  a.p2 Let FQP = :  tan (90  ) = p cot  = p 1 tan    p Also: a. p 2  a M FP  2.a . p p2  1 2. p Using the angle between two lines formula: p2  1 p 2. p tan   p2 1 1 2 1 tan   p  tan  = tan   FPQ = PQF =

Luke Cole

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Parabola I

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The Tangents at the Extremities of a Focal Chord Intersect at Right Angles on the Directrix: i.e.

QZP = 90

&

Q

Lies on the Directrix

F

P

F = Focal Chord = (0, a) Q = Tangent = (2.a.q, a.q2) P = Tangent = (2.a.p, a.p2) Directrix: y =  a

Z

Proof: Since PQ is a focal chord, p.q = 1; Tangent at P has gradient m1 = p; Tangent at Q has gradient m2 = q: p.q =  1 Use perpendicular general form:  m1.m2 =  1  Tangents are perpendicular Tangents intersect at [4]:  [a(p + q), a.p.q] So: y = a.p.q …(1) Sub p.q = 1 into (1): y=a This is the equation of the directrix  Tangents intersect on the directrix

Luke Cole

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