Week 6 – SS2 Second Term Further Mathematics Notes

 WEEK SIX
CONIC SECTIONS: PARABOLA,ELLIPSE AND HYPERBOLA
THE PARABOLA
The parabola is a locus of points, equidistant from a given point, called the Focus and from a given line called the Directrix.

(Length of directrix from V, (AV) = Length of Focus from V ,(FV))
The line AB, a distance ofa, from the y axis is called the Directrix. The line AF is called the axis of symmetry.
Since
BP = FP
BP² = FP²
(x + a)² = (x-a)² + (y-0)²
    x² + 2ax + a² = x² – 2ax + a² + y²
    4ax = y²
thus, y² = 4ax is the equation of the parabola.
The line RQ which is perpendicular to AF is called the latusrectum, V is called the vertex and F the focus of theparabola.
If the vertex of the parabola is translated to a point \(x₁,y₁), the equation of the parabola becomes
(y-y₁)² = 4a(x-x₁)².
The above equation is said to be in the standard or canonicalform
Examples
1. find the focus and directrix of the parabola y² = 16x
2. write down the equation of the parabola y²– 4y-12x+40 = 0 in its canonical form and hence find i) the vertex; ii) the focus; iii) the directrix of the parabola
`    Solution
1. comparey² = 16x with y² = 4ax,
4a = 16 , a = 4
Thus the focus is (4,0) while the directrix is x = -4
2. y²– 4y-12x + 40 = 0
y²– 4y-+4-12x+40 = 0+4 …. (completing the square)
y²– 4y+4 = 12x – 36 …… (rearranging)
(y – 2)² = 12(x-3) …… (factorising)
But (y – y₁)² = 12(x-x₁)
i) hence vertex (x₁,y₁) = (3,2)
ii) since 4a = 12, a = 3 then the focus (x₁+a,y₁) = (3+3,2) = (6,2)
iii)the equation of the directrix is x = 3-3 ie x=0
note that the thedirectrix is of equal but opposite distance from the vertex with thefocus
this means the distance between the focus and the vertex = the distance between the directrixand the vertex

 
 Equation of Tangent and Normal at point (x₁,y₁) to a Parabola
1. Equation of Tangent
=
2. Equation of Tangent
= –
Example: find the equation of the tangent and normal to a parabola
i) y² = 12x at point (3,6)
ii) y² = 16x at point (1, -4)

 Solution
ii)y² = 16x , compare with y² = 4ax, thus a= 4

 equation of tangent
= ‘ =
Thus
y + 2x +2 =0 is the equation of the tangent

 equation of the normal
= – , = –
Thus
2y –x + 9 = 0 is the equation of the normal

 Evaluation
1. Find the foci and directrices of the following Parabolae
(a) y² = 32x (b) x² = 12y
2. Write the equation of the parabola, y² – 6y – 2x + 19 = 0 in the canonical form hence determine
Its vertex and focus

 THE ELLIPSE
An ellipse is the locus of a point P, moving in a plane such that the sum of its distances from two fixed points F₁ and F₂ called the foci, is a constant.

 
OV = PF =a, OP = b and OF = c
Where V is the vertex or vertices, and F is the focus or foci .
The equation of an ellipse is given by
+ = 1 or + = 1 with centre (x¹ , y¹) (a>b) major axis on x……eqn(i)

  + = 1 or + = 1 with centre (x¹ , y¹) (a>b) major axis on y……eqn(ii)

 a²= b² + c²
wherea and b are on the major and minor axis respectively.

Examples
1. Find the foci and four vertices of the ellipse + = 1
2. Write down the equation of the ellipse 25x²+ 4y²-50x-16y-59 = 0 in the canonical form and
hence find
i) the coordinates of the centre of the ellipse
ii) the four vertices of the ellipse
iii) the two foci of the ellipse

 Solution
1. + = 1
Since a>b , then
+ = +

 By inspection,
a²=25 and b²=9 thus a=+5 or -5 , b= +3 or -3 and c = + or –
c²=a²-b² , c= +4 0r -4
i) the foci f(0,c) = f₁(0,4) and f₂(0,-4)
ii) the vertices V(a,0) = V₁(3,0) and V₂(-3,0)
the co-vertices V(0,b) = V₃(o,5) and V₄(o,-5)

 2.     25x²+ 4y²-50x-16y-59 = 0
    25x²-50x + 4y²-16y = 59
25(x²-2x +1-1)+ 4(y²-16y+4-4) = 59
25(x²-2x +1) -25+ 4(y²-16y+4)-16 = 59
25(x²-2x +1) + 4(y²-4y+4) = 100
        25(x-1)²+ 4(y-2)² = 100
         + = 1    
         But + = 1
    By inspection,a = +5 or -5 and b = +2 or -2
i) the coordinates of centre = (1,2)
ii) the vertices of the vertical axis are
V(0+x₁,a+y₁ ) = V₁(1,7)
V(0+x₁,-a+y₁ ) = V₂(1,-3)
vertices of the vertical axis are
V(b+x₁ , 0 + y₁) = V₃(3,2)
V(-b+x₁ , 0+y₁) = V₄(-1,2)
iii) the two foci are
F(0+x₁,c+y₁ ) = F₁(1 , +2)
F(0+x₁,-c+y₁ ) = F₁(1 , +2)

 Equation of Tangent and Normal at (x₁ , y₁) to an Ellipse
= is the equation of the tangent
= is the equation of the normal

 Example: find the equation of the tangent and normal to the ellipse 4x² + 25y² = 100
Evaluation
1. Find the foci and vertices of the following ellipses
(a) 9x² + 10y² = 90 (b) 4y² + 5x² = 20
2. Write the equation of the ellipse, 4x² + 5y² – 24x – 20y + 36 = 0 in the canonical form hence determine
Its vertices and foci

 THE HYPERBOLA
The hyperbola is the locus of a point P, moving in a plane such that the distance from two fixed points called the foci have a constant difference

 
The equation of an ellipse is given by
– = 1 or – = 1 where = –

 The equation of a Tangent to an ellipse at point (x₁ , y₁) is given by
=

 The equation of a Normal to an ellipse at point (x₁ , y₁) is given by
=
Examples
1. Find the vertices and foci of the Hyperbola 25x² – 16y² = 400
2. Find the equation of the tangent and normal to the Hyperbola 9x² – 36y² = 36
Solution
1. 25x² – 16y² = 400
– = 1 (in the canonical form)
Since – = 1 , then
a = +4 or -4 b = +5 or -5 c = + or – hence
the vertices are V₁(4,0) and V₂(-4,0)
the foci are F() and F()

 The General Conic
A conicin general may be defined as the locus of a moving point P, such that its distance fixed
Point called the focus, and its distance from a fixed line called the directrix are in constant ratio.
This constant ratio is called the eccentricity of the conic denoted by e. for

1. a parabola        e = 1
2. an ellipse        e < 1
3. a hyperbola    e > 1

 GENERAL EVALUATION
1. Write the equation of the ellipse, x² + 3y² + 2x –24y + 46 = 0 in the canonical form hence determine. Its vertices and foci
2. Find the vertices and foci of the hyperbola 25x² – 4y² = 100
3. Find the equation of the tangent and normal to the parabola y² – 18x = 0 at point (2,6)

 READING ASSIGNMENT: New Further Mathematics Project 3 by TuttuhAdegun and Godspower5th Edition    

 WEEKEND ASSIGNMENT
1. Find the equation of the tangent to the ellipse 4x² +9 y² = 36 at point (0,-2)
A, 6y + x = 9 B, 3y + x = 9 C, 6y + 4x = 9 D, y +6 x = -9
2. Find the foci of the ellipse 4x² + 9y² = 72
A, 14 or -14 B, or – C, 5 or -5 D, or –
3. Find the equation of the normal to the parabola y²-20x = 0 at (3 , 2 )
A. 5y + x = 13 B. 3y + x = 9 C. 5y + x = 13 D. 5y + x = 13
4. Which of the following is true eccentricity e of a parabola A.e< 1 B. e> 1 C. e = 1 D. e >= 1
5. Which of the following is true eccentricity e of a hyperbola A. e < 1 B. e > 1 C. e = 1 D. e >=1

 THEORY
1. (a) Show that the points Q(6 , 2) lies on a circle x² + y² – 4x +2y -20 = 0 lies on a circle
(b) Find the equation of the tangent to the circle at the point Q
2. write the equation of the following ellipses in their canonical form and hence determine
Their foci and vertices

1. 4x² + 5y² – 24x – 20y + 36 = 0
2. 4x² + 6y² – 24x + 60y + 162 = 0