# 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 – c2 = 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.

.