PARABOLA
Parabola
is the locus of a point which moves so that its distance from a fixed point and
a fixed line are equal.
The
fixed point is called the focus of the parabola and the line is called the directrix of the parabola.
DD’
– directrix
E, E’
– end points of the latus rectum
V
– vertex of the parabola
F – focus of the parabola
VF = a
– focal distance
DD’V =
VF = a
EF = FE’
= 2a
The axis of the parabola is horizontal ( OX
).
From the above figure and by
definition of parabola :
PF = PQ
Ö ( x – a )2 + ( y – 0 )2 = ( x + a )
Square both sides
( x – a )2 + y2
= ( x + a )2
X2 – 2ax + a2 + y2 = x2 + 2ax
+ a2
X2 – x2 + a2
– a2 + y2 = 2ax + 2ax
y2 = 4ax
Standard
equation of a parabola having horizontal axis and opens to the right.
From the figure above and by
definition of parabola :
PF = PQ
Ö( x – 0)2 + ( y – a )2 = y +
a
Square both sides
X2 + y2 –
2ay + a2 = y2 + 2ay + a2
X2 + y2 – y2
+ a2 – a2 = 2ay +
2ay
X2 = 4ay
Standard
equation of a parabola with vertical axis (OY) and opens upward.
General Equation of a Parabola
1. x2 + Dx + Ey + F = 0 , E ¹ 0, with
vertical axis ( OY )
2. y2 + Dx + Ey + F = 0, D ¹ 0, with
horizontal axis (Ox )
Standard Equation of the Parabola
A.
with vertex at V( 0, 0 )
1.
y2 = 4ax ----
opens to the right
2.
y2 = – 4ax ----
opens to the left
3.
x2 = 4ay ---- opens upward
4.
x2 = – 4ay ----
opens downward
B.
with vertex at V( h, k )
1.
( y – k )2 = 4a( x – h
) ---- opens to the right
2.
( y – k )2 = – 4a( x –
h) ----
opens to the left
3.
( x – h )2 = 4a( y – k )
---- opens upward
4.
( x – h )2 = – 4a( y –
k ----
opens downward
Examples.
Draw
the graph of the following parabola.
1. 2x2
– 8y = 0 3. x2
+ 2x – 6y = 2
2. 3y2
+ 18x = 0 4. y2 + 2y – 6x = 2
5. x2
+ 4y – 4 = 0 8. y2
+ 8x + 16 = 0
6. y2
– 4x + 4 = 0 9. x2
= 4x + 4y
7. y2
+ 2x + 6y + 17 = 0 10. x2 – 2x + 2y + 7 = 0
QUADRATIC
FUNCTIONS
A function of the form y = ax2
+ bx + c, where a ¹ 0 is a quadratic function. Since the equation has
the form of the general equation of a parabola x2 + Dx + Ey + F = 0, the graph of
every quadratic function is a parabola with a vertical axis. It is concave
upward if a is positive or concave downward
if a is negative.
At the vertex of the parabola, x = – b/2a. This value of x gives the
function its minimum value if a is positive or concave downward if a is
negative. The equation of the axis of the parabola is x = – b/2a.
As x increases, many functions increase
to a maximum value, decreasing thereafter, or decrease to a minimum and begin
to increase. In such cases, it is a problem of prime importance to determine
the maximum or minimum value. To solve problems for functions in general,
differential calculus is required. For the quadratic function it may evidently
be solve by merely finding the vertex of the parabola, since the ordinate of
the vertex is the greatest or least ( algebraically ) of all the ordinates,
according as the parabola opens downward or upward.
STEPS
IN GRAPHING A QUADRATIC FUNCTION
1. Compute x = – b/2a, and then the
corresponding value of y
to obtain the coordinates of the vertex
of the parabola.
2. If the value of x = – b/2a is convenient, form a table by using
pairs of values of x where each pair, the
values are
equidistant from the vertex on either
side. The value of y
corresponding to any pair will be equal.
3. If the value of x = – b/2a
is inconvenient, form the table by
choosing
values of x arbitrarily on either side of
x = – b/2a.
Example. Determine the critical points whether it is
maximum or minimum.
1. y
= x2 – 10x + 25
2. y = 4x
– x2
3. y = 2x2
– 10x + 3
4. y = 15x – 3x2
Application
problems.
Use
quadratic function to evaluate the following problems.
1. Find the dimensions of the largest
rectangular lot that can be
enclosed by 100 m of fencing material.
2. A rectangular field is to be fenced off
along the bank of a
river. If no fence is needed along the
river, what is the shape
of the field requiring 400 m of fence if the
area is to be
maximum?
1. Solution
Let x and y be the dimensions (length and width)
of the lot
Area :
A = xy
Perimeter
: 2x + 2y = 100
x + y = 50
y = 50 – x
x(50 – x ) = A
50x – x2 = A
Standardize the
equation :
X2 – 50x = – A
x
= – b/2a
= – (–50) / 2 ( 1 )
x =
25 m
y = 50 – x
y = 50 – 25
y = 25
m
Alternately :
By completing the square :
x2 – 50x + 625 = – A + 625
( x – 25 )2 = – ( A – 625 )
x
– 25 = 0 | A – 625 = 0
x = 25 m
| A = 625 m2
y
= 50 – 25
= 25 m
The dimensions are 25 m by 25
m.
The maximum area is 625 m2
B. Use
quadratic functions to solve the problems.
1. A rectangular field is to be enclosed,
and divided into three
lots by fences parallel to one of the
sides. Determine the
dimensions of the largest field that can
be enclosed by
2,400 m of fencing material.
2. A
rectangular lot is to be fenced off
along a highway. If the
fence along the highway cost P 150
per meter, and on the
other sides P 100
per meter, determine the dimensions of
the largest lot that can be
enclosed for a budget of
3. A
rectangular field is to be fenced off along the bank of a
river. If no fence is needed along the
river, what is the
shape of the field requiring 400 m of fence if the area is
to
be maximum?
4. A triangle has a base of 12 cm
and an altitude of 8 cm.
Find the area of the largest rectangle that
can be inscribed
in
the triangle so that the base of the rectangle falls on the
base of the triangle.
5. A right
triangle has a hypotenuse of 30 cm. Show that the
area is maximum when the triangle is
isosceles.
