Conic Sections

 

FIRST TERM

WEEK 5

SUBJECT: FURTHER MATHEMATICS

CLASS: SS 3

TOPIC: Conic Sections

CONTENT:

1. Equations of parabola, Eclipse, Hyperbola in rectangular Cartesian coordinates
2. Parametric equations

Sub-Topic (i): Equations of parabola, ellipse, hyperbola in rectangular coordinates

The path (or locus) of a point which moves so that its distance from a fixed point is in constant ratio to its distance from a fixed line is called a conic section or a conic.

The fixed point F is called the focus, the fixed line is called the directrix, and the constant ratio is called eccentricity, usually denoted by e.

F (Fixed point)

Fixed line

M

P

Thus, if F is a point on the locus (as shown in the diagram above), M is the foot of the perpendicular from P to the directrix and if , then the locus of F is a conic. If e<1, the conic is an ellipse. If e = 1, the conic is a parabola. If e>1, the conic is a hyperbola.

Parabola: a parabola is the locus of points in a plane which is equidistant from a fixed line and a fixed point. Let L be the fixed line and F the fixed point as shown below. Through F draw the x-axis perpendicular to the fixed line of distance 20 units from F. From the definition of the parabola, the curve must cross the x-axis at a point 0, midway between F and L. Then draw the y-axis through 0. The co-ordinates of F are (a, 0).

a

(-a, 0)

M(-a, y)

x

s

P(x,y)

F(a,0)

R

V

y

If p(x,y) is any point on the parabola such that … (i)

then (PF)² = (x-a)² + y² (from Pythagoras rule)

and PM = X+A; but (from equation 1)

PM = PM; so, (PF)² = (PM)² (by squaring both side)

⇒ (x-a)² + y² = (x+a)²

or x² – 2ax + a² + y² = x² + 2ax + a²

therefore, the equation of a parabola is y² = 4ax

the line segment through the focus and perpendicular to the axis of symmetry, and with end points S and R on the parabola, is called the lotus rectum. The point V is called the vertex of the parabola.

Special Cases;

y²=-4ax

F(-a,0)

x

y

(b)

y²=4ax

F(a,0)

R

x

y

(a)

x²=-4by

F(0, -b)

x

y

(d)

directrix

x²=4by

F(-a,0)

x

y

(c)

directrix

If the vertex of the parabola y² =4ax is translated to the point (x₁,y₁), the equation of the corresponding parabola becomes (y-y₁)² = 4a(x-x₁)and if the vertex of the parabola becomes (x-x₁)² = 4a(y-y₁). The above equations are said to be in their standard or canonical form.

Examples:

(1) Find the foci and directices of these parabolas:

1. y² = 32x[mediator_tech]
2. x² = 12y

Solution:

1. by comparing y² ₌ 32xwith y² =4ax,

a = 8.

Hence, the focus is (4,0) while the directrix is x = -4

1. by comparing x² = 12y with x² = 4by

b = 3.

Hence, the focus is (0,3) while the directrix is y = -3.

(2) Write the equation of the following parabola in their canonical forms and hence determine their vertices, foci and directrices:

(a) y²-6y-2x+19 = 0

(b) x²+4x+4y+16 = 0

Solution:

1. y²-6y+9-2x+10 = 0 (complete the square for y)

y²-6y+9=2x-10

(y-3)² = 2(x-5)

Comparing our solution with the general with the general canonical form, i.e.

(y-y₁)² = 4a (x-x₁)

⇒ y₁ = 3, x₁ =5, 4a = 2 a = ½

Hence, the vertex is (5,3), the focus (5 + ½, 3) = (11/2, 3) and the directrix is x = 5 – ½ = 4½ = 9/2.

1. x²+4x+4y+16 = 0

x²+4x+4+4y+12 = 0

x²+4x+4 = -4y-12

(x+2)² = -4(y+3)

Comparing our solution with the general canonical form,

⇒ x₁ = -2, y₁ = -3,

(x-x₁)² = 4b(y-y₁)

⇒ x₁ = -2, y₁ =-3,

Hence, the vertex is (-2, -3), the focus (-2, -3+(-1)) = (-2, -4) and the directrix is y = -3-(-1) = -3+1 = -2

Equations of the tangent and normal to the parabola

1. equation of the tangent to y² = 4ax at point (x₁,y₁) by differentiating y² = 4ax implicitly, we obtain

At (x₁,y₁)_,

y₁ ( = 2a (

y₁ = 2ax – 2ax,

Since (x₁,y₁) is on y² = 4ax, ⇒ y₁² = 4ax₁

Thus, yy₁-4ax = 2ax – 2ax₁

yy₁=2ax-2ax₁ + 4ax₁

yy₁ = 2a(x+x₁)

Equation of the normal to y₁ = 4axat the point (x₁,y₁)

⇒ (x-)

slope of the normal is , since the normal is perpendicular to tangent, and m₁m₂ = -1, where m₁ and m₂ are the slopes of the normal and the tangent at (x₁, y₁)

⇒) or

Example (3): Find the equation of the tangent and normal to the parabola y² = 12x at one end of its latus rectum.

Solution:

y² = 12x compared to y² = 12x⇒ a = 2

One end of its latus rectum is L (3,6)

Tangent at (x₁,y₁) is yy₁ = 2a(x+x₁)

Here, (x₁,y₁) ₌ (3,6) and a = 3

Tangent at L is y(6) = 6(x+3)i.e. y = x+3

Normal at (x₁,y₁) is =

So normal at L is = i.e. y=-x+9

Example (4):

Find the equations of the tangent and normal at (1,2) on the parabola x²+x-2y+2 = 0

Solution:

Differentiate x²+x-2y+2 = 0 with respect to x (w.r.t.x) to get the slope(m)

⇒ ,

At (1,2), = ⇒ Equation of the tangent is = m (x-x₁)

i.e. y-2= or 2y-4 = 3 (x-1)⇒ 2y – 4 = 3x – 3, 3x-2y+1 = 0

Equation of the normal at (1,2) is

y – y₁ = – (x-1) or 3y – 6 = -2(x-1)

. 2x + 3y – 8 = 0

THE ELLIPSE: An ellipse is the points, the sum of whose distance from two fixed points is constant.

If the two fixed points, called foci, are taken at F₁(-c,0) and F₂(c,0)

y

x =

x = –

(0,b)V₃

P(x,y)

x

V₂ (-a,0)

F₁ (-c,0)

0

(a,0)V₁

F₂ (c,0)

(0,-b)V₄

Directrix

Directrix

Then from the definition,

PF₁ + PF₂ = 2a⇒

Squaring both sides and collecting terms, we have

• y² = 4a² – 4a + +y²

x² + 2xc + c² + y² = 4a² – 4a + x² – 2xc + c² + y²

4xc – 4a² = -4a

Divide through by -4a

=

• a =

Square both sides

⇒ – y² = x² – a² + c² + a² – c^{2 =} x² + y² a² – c² = y² + x² –  ; a² – c² = y² + x² ()

Multiply through by[mediator_tech]

= ()x

1 = + … (i)

Hence, the four vertices are V₁( 0), V₂( 0), V₃(0,3) and V₄(0,-3)

Since c² = a² – b²

c = ± 1; Hence,

F₁(-1, 0) and F₂ (1,0)

1. 5y² + 6x² = 30

To compare 5y² + 6x² = 30 with + = 1

First, divide through by 30

⇒ + = ⇒= 1⇒b² = 6, b = ±

Hence, the four vertices are V₁(), V₂(), V₃(0) and V₄(,0)

Since c² = a² – b², ⇒ c² = 6- 5 = = ±1 F₁(0, -1) and F₂ (0, 1)

(2) Write the equation of the ellipse 4x²+5xy²-24x-20y+36 = 0 in the canonical form and hence, determine:

1. the coordinates of the centre of the ellipse;
2. The four vertices of the ellipse;
3. The two foci of the ellipse

Solution:

4x²+5xy²-24x-20y+36 = 0

Rearrange :

4x²+5xy²-24x-20y = -36

4(x²-6x) + 5(y²-4y) = -36

4(x²-6x+9-9) + 5(y²-4y + 4-4) = -36

(Complete each square in x and y and subtract the value added for completing the square)

4(x²-6x+9) – 36 + 5(y-2)² – 20 = -36

4(x-3)² – 36 + 5(y-2)² -20 = -36

4(x-3)² + 5(y-2)² = -36 + 36 + 20

4(x-3)² + 5(y-2)² =20

Divide through by 20 so we can compare with = 1

⇒ + =⇒ + = 1

Hence,

1. The coordinate of the centre = (6,2)
2. Comparing + = 1 with + = 1

⇒ b² = 5, b = ±, a² = 4, a = ±2

Hence, the vertices on the vertical axes are

V₁(0+x₁, a + y₁) = V₁(3,4),

V₂ (0+x, -a + y) = V₂ (3,0),

V₃(b+x₁, 0+y₁) = V₃(,

V₄(-b+x₁, 0+y₁) = V₄(,

The of the tangent, and at normal (x₁,y₁) to the Ellipse

By differentiating the equation of ellipse i.e. + = 1 implicitly and substituting for , we obtain + = 1 as equation of tangent.

For equation of the normal at (x₁, y₁), since m1m2 = -1 for perpendicular lines, it implies that by dividing -1 by got from implicit differentiation of equation of ellipse and substituting x₁ and y₁, we obtain.

=

Example (4): Find the equations of the tangent and normal to the ellipse at x²+4y² = 4 at (4,1)

Solution:

Rewriting

x²+4y² = 4, we have

• = 1⇒ a² = 4 , a = 2; b² = 1, b= 1

Equation of tangent at (x₁,y₁) is + = 1

Hence, the equation of the tangent at (4, 1) is

• = 1 , ⇒ x + y = 1

Equation of the normal at (x₁, y₁) is

= ⇒ =

y – 1 = x – 4 ; x – y = –3

The hyperbola: a hyperbola is the locus of a point P, moving in a plane such that its distances from two fixed points called foci have a constant difference. For a hyperbola, the eccentricity (e) > 1.

y

P(x,y)

x

F₁ (-c,0)

0

F₂ (c, 0)

V₁ (a, 0)

(-c,0)V₂

From the definition of a hyperbola and following the diagram above, we have PF₁ – PF₂ = 2a[mediator_tech]

⇒ – = 2a

By simplifying and substituting b² = c² – a² we obtain – = 1

If the centre of the hyperbola is translated to the point (x₁, y₁), the equation of the hyperbola in canonical form becomes

• = 1

The points V₁(a,0) and V₂(-a,0) are the vertices of the hyperbola. The points F₁(c,0) and F₂(-c,0) are the foci of the hyperbola.

ASYMTOTES OF THE CURVE

The left hand side of the equation of a hyperbola is a difference of two squares, i.e.

• = 1 ⇒ = 1⇒⇒ =

Thus, =

Consider the R.H.S. of the equation; when we divide the numerator and denominator by x, we have

= ; as y → , → 0

Also, ; as y → , → 0

So as x and y approach infinity the expression on the R.H.S of the equation = approaches zero. Hence, the expression of the L.H.S must also approach zero as x and y approach infinity i.e.

→ 0

The straight line = or y = is called an oblique asymptote. The curve is symmetrical about the axis, so the line y = is also another oblique asymptote.

Equation of the tangent and normal at the point (x₁, y₁) is + = 1 and the equation of the normal is = 1 or a²xy₁ + b²x₁y = (a² + b²)x₁y₁

Examples (5): find the vertices and foci of the parabola 25×2 – 4y2 = 100

Solution:

⇒

Comparing with the canonical form, i.e.

We have a² = 4, ⇒ a = ±2 and b² = 25, ⇒ b = ±5

Since b² = c² – a², ⇒ c² – a², ⇒ c² = b² + a² = 25 + 4 = 29 a = ±

The vertices are V₁(2,0) and V₂(-2,0) and the foci F₁(-, 0) and F₂(, 0)

Example (6): obtain the equation of the hyperbola whose foci are (6,4) and (-4, 4) and the electricity is 2.

Solution:

Let F₁, F₂ and C lie on the line y = 4. The coordinates of the centre the mid-point of the line joining the vertices which are

The equation of the hyperbola is therefore

Shifting the origin to (1,4), we put x = x₁+1, y = y₁+4

Then the equation becomes

The distance between the two foci is 2ae (since e = , and this is equal to the actual distances 10.

2ae = 10, ⇒ 2a =

Since b² =c² – a²⇒ b² =a² e² – a²

b² = a² (e² -1) =

Hence the equation of the hyperbola is

⇒ – = 1

Multiplying through by 75, we have

12(x-1)² – 4(y-4)² =1

12(x²-2x+1)-4(y²-8y+16) =1

12x²-24x+12-4y²+32y-64-1=0

12x²-4y²-24x+32y-53=0

EVALUATION

1. Write down the equation of the ellipse 2x²+5y²+8x+10y+3=0
2. Write the equation of the parabola y²+4y+4x+16=0 in its canonical form and determine the vertices, foci and directrix.
3. Find the equations of the tangent and the normal to the hyperbola 4x²-5y²=20 (.

SUB-TOPIC 2:

Parametric Equations of parabola, Ellipse and hyperbola so far considered are of the form y =f(x), where a direct relationship between x and y is given. This form is called the Cartesian equation.

However, it is more convenient sometimes to express both x and y in terms of a third variable, called a parameter, that is, x=f(t), y=g(t)

Each value of the parameter ‘t’ gives a value of x and a value of y.

Parametric Equations of the straight line

To find the parametric equations of a straight line passing through a given point (x₁,y₁) and inclined to the x-axis at an angle θ with 0x. Let the line be considered as the locus of a moving point p(x,y). Let PQ = r.

Draw PM and QN perpendicular to the x-axis and QL perpendicular to PM. Then, = cosθ and

Q(x,y)

θ

x₁

x-x₁

M

x

N

0

y

P(x,y)

A

y-y₁

y₁

r

L

or = r and = r

or

⇒ x₁ + r cos, y=y₁ + sin

These are parametric equations of a straight line through the paint (x₁,y₁) with inclination . It follows that the parametric coordinates are (x₁ + r cos, y=y₁ + sin).

Parametric equations of a circle, x² + y² = a²

Let the equation of a circle be x² + y² = a². Let p(c,y) be any point on the circle and PM be the perpendicular from p to the x-axis. Let OP, which is equal to a in length, make an angle with OX.

P(x,y)

x

y

Then x = acos[mediator_tech]

These are the parametric equations of the circle x² + y² = a² where θ is the parameter. It also follows that the parametric coordinates are (acosθ, asinθ). If the equation of the circle is in general form, i.e. (x-h)² + (y-k)² =a², then the parametric equations are (h+acosθ, K+asin θ).

Parametric equations of the parabola y²=4ax

The y²=4ax when divided through by 2ay gives

Then, y =2at and x =

Hence, the parametric equations of the parabola y²=4ax are x=at² and y=2at, where t is the parameter.

y=2at

x

y

x=at²

Since the point (at², 2at) satisfies the equation y²=4ax, therefore the parametric coordinates of any point on the parabola are (at², 2at).

Tangents and Normals of Parametric Equations

We use the relationship (composite rule) ==

To find the gradient of a curve in terms of the parameter and hence the equations of the tangent and normal.

Examples:

1. Find the parametric equations of the line through (3,-2) having inclination equal to 135⁰.

Solution:

Parametric coordinate are (x + r cos, y + sin)

⇒ x = 3 + rcos θ 135, y = -2+rsin135

⇒ x = 3 – rcos 45⁰, y = -2+rsin45

⇒ x = 3 – , y = -2+

1. Find the parametric equations of the circle (1,5) and radius 3.

Solution:

Parametric equations of at circle at (1,5) and r = 3 given by x = h+acosθ, y=k+asin θ

⇒ x = 1 + 3cosθ, y = 5+asinθ

1. Find the equations of the tangent and normal to the curve x =4t-1, y=2t² at the point where t = 1

Solution: the gradient of the tangent is given by

= =

The equation of the tangent at the point (4t-1, 2t²) will be (y-2t²) = t[x-(4t-1)] i.e. from y-y₁ = m(x-x₁), equation of straight line

⇒ y-2 = (x-3) ⇒ y = x-1

The gradient of the normal is -1/t

Hence, the equation of the normal at (4t-1, 2t²) is

y-2t² =-1/t(x-4t+1)

when t=1, y-2 = -1(x-4+1) ; y = 3-x ;y=5-x

EVALUATION

1. Find the equation of the line tangent to the given curve at the given point:
2. X=2t-1, y=4t²-2t; t=1 (b). X=3cosθ, y=2sinθ; θ=π/4
3. Find the Equation of tangent on the curve given by x = , y = , at the point where t = 1.

GENERAL EVALUATION

1. Find the equation of the parabola with focus (3,0) and directrix 1. (a) y²=3x (b) y²=4x (c) y²=6x (d) y²=12x
2. Which of the following does the equation 4x²-4y² = a represents?
3. Find the parametric equation passing through (-1,4), and inclined at angle θ =60⁰ (a) x = -1-, y= (b) x = 1+, y=-5+2r (c) x=1+, y =1- (d) x=-1+, y =5-

THEORY

1. Find the eccentricity of the ellipse 2x²+4y² = 4
2. Show that if the line Ax +By +C = 0 must be a tangent to the hyperbola , then the condition A²a²-B²b² = C² must be satisfied.

.