# Conic Sections

**FIRST TERM**

**WEEK 5 **

**SUBJECT:** FURTHER MATHEMATICS

**CLASS**: SS 3

**TOPIC:** Conic Sections

**CONTENT: **

- Equations of parabola, Eclipse, Hyperbola in rectangular Cartesian coordinates
- 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} + y^{2} (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} + y^{2} = (x+a)^{2}

or x^{2} – 2ax + a^{2} + y^{2} = x^{2} + 2ax + a^{2}

therefore, the equation of a parabola is y^{2} = 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;

y^{2}=-4ax

F(-a,0)

x

y

(b)

y^{2}=4ax

F(a,0)

R

x

y

(a)

x^{2}=-4by

F(0, -b)

x

y

(d)

directrix

x^{2}=4by

F(-a,0)

x

y

(c)

directrix

If the vertex of the parabola y^{2} =4ax is translated to the point (x_{1},y_{1}), the equation of the corresponding parabola becomes (y-y_{1})^{2} = 4a(x-x_{1})and if the vertex of the parabola becomes (x-x_{1})^{2} = 4a(y-y_{1}). The above equations are said to be in their standard or canonical form.

**Examples: **

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

- y
^{2}= 32x[mediator_tech] - x
^{2}= 12y

Solution:

- by comparing y
^{2}_{ = }32xwith y^{2}=4ax,

a = 8.

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

- by comparing x
^{2}= 12y with x^{2}= 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) y^{2}-6y-2x+19 = 0

(b) x^{2}+4x+4y+16 = 0

Solution:

- y
^{2}-6y+9-2x+10 = 0 (complete the square for y)

y^{2}-6y+9=2x-10

(y-3)^{2} = 2(x-5)

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

(y-y_{1})^{2} = 4a (x-x_{1})

⇒ y_{1} = 3, x_{1} =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.

- x
^{2}+4x+4y+16 = 0

x^{2}+4x+4+4y+12 = 0

x^{2}+4x+4 = -4y-12

(x+2)^{2} = -4(y+3)

Comparing our solution with the general canonical form,

⇒ x_{1} = -2, y_{1} = -3,

(x-x_{1})^{2} = 4b(y-y_{1})

⇒ x_{1} = -2, y_{1} =-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

- equation of the tangent to y
^{2}= 4ax at point (x_{1},y_{1}) by differentiating y^{2}= 4ax implicitly, we obtain

At (x_{1},y_{1})_{, }

y_{1} ( = 2a (

y_{1} = 2ax – 2ax,

Since (x_{1},y_{1}) is on y^{2} = 4ax, ⇒ y_{1}^{2} = 4ax_{1}

Thus, yy_{1}-4ax = 2ax – 2ax_{1}

yy_{1}=2ax-2ax_{1} + 4ax_{1}

yy_{1} = 2a(x+x_{1})

Equation of the normal to y_{1} = 4axat the point (x_{1},y_{1})

⇒ (x-)

slope of the normal is , since the normal is perpendicular to tangent, and m_{1}m_{2} = -1, where m_{1} and m_{2} are the slopes of the normal and the tangent at (x_{1}, y_{1})

⇒) or

**Example (3):** Find the equation of the tangent and normal to the parabola y^{2} = 12x at one end of its latus rectum.

Solution:

y^{2} = 12x compared to y^{2} = 12x⇒ a = 2

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

Tangent at (x_{1},y_{1}) is yy_{1} = 2a(x+x_{1})

Here, (x_{1},y_{1})_{ = }(3,6) and a = 3

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

Normal at (x_{1},y_{1}) 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 x^{2}+x-2y+2 = 0

*Solution: *

Differentiate x^{2}+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-x_{1})

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 – y_{1 } = – (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 F_{1}(-c,0) and F_{2}(c,0)

y

x =

x = –

(0,b)V_{3}

P(x,y)

x

V_{2} (-a,0)

F_{1} (-c,0)

0

(a,0)V_{1}

F_{2} (c,0)

(0,-b)V_{4}

Directrix

Directrix

Then from the definition,

PF_{1} + PF_{2} = 2a⇒

Squaring both sides and collecting terms, we have

- y
^{2}= 4a^{2}– 4a + +y^{2}

~~x~~^{2} + 2xc + ~~c~~^{2} + ~~y~~^{2} = 4a^{2} – 4a + ~~x~~^{2} – 2xc + ~~c~~^{2} + ~~y~~^{2}

4xc – 4a^{2} = -4a

Divide through by -4a

=

- a =

Square both sides

⇒ – y^{2} = x^{2} – a^{2} + c^{2} + a^{2} – c^{2 = }x^{2} + y^{2} a^{2} – c^{2} = y^{2} + x^{2} – ; a^{2} – c^{2 }= y^{2} + x^{2} ()

Multiply through by[mediator_tech]

= ()x

1 = + … (i)

Hence, the four vertices are V_{1}( 0), V_{2}( 0), V_{3}(0,3) and V_{4}(0,-3)

Since c^{2} = a^{2} – b^{2}

c = ± 1; Hence,

F_{1}(-1, 0) and F_{2} (1,0)

- 5y
^{2}+ 6x^{2}= 30

To compare 5y^{2} + 6x^{2} = 30 with + = 1

First, divide through by 30

⇒ + = ⇒= 1⇒b^{2} = 6, b = ±

Hence, the four vertices are V_{1}(), V_{2}(), V_{3}(0) and V_{4}(,0)

Since c^{2} = a^{2} – b^{2}, ⇒ c^{2} = 6- 5 = = ±1 F_{1}(0, -1) and F_{2} (0, 1)

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

- the coordinates of the centre of the ellipse;
- The four vertices of the ellipse;
- The two foci of the ellipse

*Solution:*

4x^{2}+5xy^{2}-24x-20y+36 = 0

*Rearrange : *

4x^{2}+5xy^{2}-24x-20y = -36

4(x^{2}-6x) + 5(y^{2}-4y) = -36

4(x^{2}-6x+9-9) + 5(y^{2}-4y + 4-4) = -36

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

4(x^{2}-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,

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

⇒ b^{2} = 5, b = ±, a^{2} = 4, a = ±2

Hence, the vertices on the vertical axes are

V_{1}(0+x_{1}, a + y_{1}) = V_{1}(3,4),

V_{2} (0+x, -a + y) = V_{2} (3,0),

V_{3}(b+x_{1}, 0+y_{1}) = V_{3}(,

V_{4}(-b+x_{1}, 0+y_{1}) = V_{4}(,

The of the tangent, and at normal (x_{1},y_{1}) 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 (x_{1}, y_{1}), since m1m2 = -1 for perpendicular lines, it implies that by dividing -1 by got from implicit differentiation of equation of ellipse and substituting x_{1} and y_{1}, we obtain.

=

** Example (4):** Find the equations of the tangent and normal to the ellipse at x

^{2}+4y

^{2 }= 4 at (4,1)

*Solution: *

Rewriting

x^{2}+4y^{2 }= 4, we have

- = 1⇒ a
^{2}= 4 , a = 2; b^{2}= 1, b= 1

Equation of tangent at (x_{1},y_{1}) is + = 1

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

- = 1 , ⇒ x + y = 1

Equation of the normal at (x_{1}, y_{1}) 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

F_{1} (-c,0)

0

F_{2} (c, 0)

V_{1} (a, 0)

(-c,0)V_{2}

From the definition of a hyperbola and following the diagram above, we have PF_{1} – PF_{2} = 2a[mediator_tech]

⇒ – = 2a

By simplifying and substituting b^{2} = c^{2 }– a^{2} we obtain – = 1

If the centre of the hyperbola is translated to the point (x_{1}, y_{1}), the equation of the hyperbola in canonical form becomes

- = 1

The points V_{1}(a,0) and V_{2}(-a,0) are the vertices of the hyperbola. The points F_{1}(c,0) and F_{2}(-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 (x_{1}, y_{1}) is + = 1 and the equation of the normal is = 1 or a^{2}xy_{1} + b^{2}x_{1}y = (a^{2} + b^{2})x_{1}y_{1}

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

*Solution: *

*⇒*

Comparing with the canonical form, i.e.

We have a^{2} = 4, ⇒ a = ±2 and b^{2} = 25, ⇒ b = ±5

Since b^{2} = c^{2} – a^{2}, ⇒ c^{2} – a^{2}, ⇒ c^{2} = b^{2} + a^{2} = 25 + 4 = 29 a = ±

The vertices are V_{1}(2,0) and V_{2}(-2,0) and the foci F_{1}(-, 0) and F_{2}(, 0)

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

**Solution: **

Let F_{1}, F_{2} 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 = x_{1}+1, y = y_{1}+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 b^{2} =c^{2 }– a^{2}⇒ b^{2} =a^{2} e^{2 }– a^{2}

b^{2} = a^{2} (e^{2} -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(x^{2}-2x+1)-4(y^{2}-8y+16) =1

12x^{2}-24x+12-4y^{2}+32y-64-1=0

12x^{2}-4y^{2}-24x+32y-53=0

**EVALUATION **

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

x_{1}

x-x_{1}

M

x

N

0

y

P(x,y)

A

y-y_{1}

y_{1}

r

L

or = r and = r

or

⇒ x_{1} + r cos, y=y_{1} + sin

These are parametric equations of a straight line through the paint (x_{1},y_{1}) with inclination . It follows that the parametric coordinates are (x_{1} + r cos, y=y_{1} + sin).

**Parametric equations of a circle, x ^{2} + y^{2 }= a^{2}**

Let the equation of a circle be x^{2} + y^{2 }= a^{2}. 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 x^{2} + y^{2 }= a^{2} 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} =a^{2}, then the parametric equations are (h+acosθ, K+asin θ).

**Parametric equations of the parabola y ^{2}=4ax**

The y^{2}=4ax when divided through by 2ay gives

Then, y =2at and x =

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

y=2at

x

y

x=at^{2}

Since the point (at^{2}, 2at) satisfies the equation y^{2}=4ax, therefore the parametric coordinates of any point on the parabola are (at^{2}, 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: **

- Find the parametric equations of the line through (3,-2) having inclination equal to 135
^{0}.

**Solution: **

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

⇒ x = 3 + rcos θ 135, y = -2+rsin135

⇒ x = 3 – rcos 45^{0}, y = -2+rsin45

⇒ x = 3 – , y = -2+

- 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

- Find the equations of the tangent and normal to the curve x =4t-1, y=2t
^{2}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, 2t^{2}) will be (y-2t^{2}) = t[x-(4t-1)] i.e. from y-y_{1} = m(x-x_{1}), 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, 2t^{2}) is

y-2t^{2} =-1/t(x-4t+1)

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

**EVALUATION**

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

**GENERAL EVALUATION **

- Find the equation of the parabola with focus (3,0) and directrix 1. (a) y
^{2}=3x (b) y^{2}=4x (c) y^{2}=6x (d) y^{2}=12x - Which of the following does the equation 4x
^{2}-4y^{2}= a represents? - Find the parametric equation passing through (-1,4), and inclined at angle θ =60
^{0 }(a) x = -1-, y= (b) x = 1+, y=-5+2r (c) x=1+, y =1- (d) x=-1+, y =5-

**THEORY **

- Find the eccentricity of the ellipse 2x
^{2}+4y^{2}= 4 - Show that if the line Ax +By +C = 0 must be a tangent to the hyperbola , then the condition A
^{2}a^{2}-B^{2}b^{2}= C^{2}must be satisfied.

.