DISTANCE AND MIDPOINT FORMULAS; PARABOLAS

In this unit, you will begin to investigate the conic sections. The conics are curves formed by the intersection of a plane and a double-napped cone. There are four types of conic sections: a parabola, a circle, an ellipse, and a hyperbola. A diagram of a double-napped cone is given below. A double napped-cone has two cones with the points touching. Notice how slicing the double-napped cone at different angles produces different conic sections. In this unit you will find the distance between two points, the coordinates of the midpoint of a line segment, and the equation of a parabola.

Parabola Circle Ellipse Hyperbola

Distance Formula

Midpoint Formula

Parabolas

Distance Formula

The distance between two points, A( x1,y 1 ) and B( x22, y ), on a coordinate plane is as follows:

22 dxxyy()()21 21 *the order of x and y does not matter.

Example #1: Find the distance between point S which is located at (3, 5) and point T which is located at (–4, –2).

d (4 3)22 (2 5)

d (7)22  (7)

d 49 49

d  98

d 49 2

d  72

The distance between S(3, 5) and T(–4, –2) is 72.

Midpoint Formula

The coordinates of the midpoint, M, between two points A(,x11y ) and B (,x22y ) can be found using the following formula:

x1212 xy y M , 22

Example #1: Find the coordinates of the midpoint between point P which is located at (–3, 2) and point Q which is located at (5, –2).

352(2)  M  , 22 20 M  , 22 MPQ(1, 0) The coordinates of the midpoint between and are (1,0).

Center, Circumference, and Area of a Circle in the Coordinate Plane

The midpoint formula can be used to find the center location of a circle on a coordinate plane; and then, by using the distance formula, you can find the area and the circumference of the circle.

Example #2: The diameter of a circle has endpoints located at P(8, –3) and Q(–2, 1)

a.) The center location of the circle is found by using the midpoint formula.

8(2)31  C  , 22 62 C  , 22 C (3, 1)

b.) To find the area of the circle  r 2 we will have to find the distance between the center (3, –1) and either point P(8, –3) or Q(–2, 1) because this will give us the radius. Let’s use point Q.

r (2 3)22 (1(1))

r (5)22  2

r 25 4

r  29 This is the radius of our circle. c.) To find the area of the circle use the formula  r 2 .

A   (29)2

A  29 d.) To find the circumference of the circle use the formula 2r .

C  229

Parabolas

Parabolas That Open Up or Down

A parabola is the set of all points in a plane that are the same distance from a given point called the focus and a given line called the directrix.

y

focus (5, 3)

(x, y) some point

x

(x, –1) directrix y = –1

The distance between the focus (5, 3) and a point on the parabola (x, y) is equal to the distance from the same point (x, y) and a point on the directrix (x, –1). Basically this is saying the blue lines have the same length.

This information can be used to find the equation of the parabola; which, if you remember, is y ax() h2 k. We will now find the equation of the example:

Given: focus (5, 3) and directrix y = –1

distance between (x, y) and (5, 3) = distance between (x, y) and (x, –1)

(5)(3)(xy22 xxy  )((1)) 2 2

-square both sides to eliminate the radicals

(5)(3)0(1)xy22 y 2

-Since there are two quantities that contain the variable y, we want to expand these quantities so we can solve the equation (expand these quantities by using the FOIL method).

(5)(3)(3)(1)(1)xyyyy2

(5)xyyyy22 69 2 21

-cancel out the y 2 terms and solve for y

(5)88x 2 y

-divide all terms by 8 11 (5)1xyyx22 or  (5)1  88

-This is the equation of our parabola in the example illustration.

Up to this point we have seen parabolas that open up or down that have a standard equation of y ax() h2 k, where the vertex is located at (h, k) and the axis of symmetry is x = h.

Parabolas That Open Right or Left

We are now going to look at parabolas that open left or right. The standard form of a parabola that opens left or right is:

x ay() k2 h, where the vertex is located at (h, k) and the axis of symmetry is y = k

if “a” is positive, the parabola opens right

if “a” is negative, the parabola opens left

Example #1: Given the equation 5426xy 2  y , we are going to put it in standard form:

1.) “complete the square” with the “y” terms, take half of 4 and square it, add this to both sides.

5(4)26xyy442   

2.) factor yy2 44 into a perfect square

54(2)26xy 2 

3.) subtract 4 from both sides

5xy ( 2)2  30

4.) divide everything by 5 to solve for x.

1 xy(2)62 5

The vertex is located at (–6, –2) and the axis of symmetry is y = –2

To view the parabola on a graphing calculator, solve the equation for y.

5xy ( 2)2  30

530(2)xy2

5302xy

2530 x y

yx2530  or

yx2530 

*Note: Due to the limitations of the graphing calculator, two equations must be entered to see the whole graph.

y 2

The vertex is located at (–6, –2) and the axis of symmetry is y = –2