Section 9-4
Polar form of linear equation
`
The polar form of a linear equation, where
p is the length of the normal and φ is the positive angle between the positive x‐axis and the normal, is
` P=r cos (θ‐φ)
1. Put the equation in standard form.
2. Convert the equation to normal form
3. Substitute the values for p and φ into the polar form
Write 12 x – 5y = -13 in polar form.
` Standard form 12 x -5y +13 = 0
A2 + B 2
` Normal form:
Find value of ±
144 + 25
=
` Since C is positive we use -13.
=
± 13
` When C is negative we use the positive form )
− 12
` So the normal form of the equation is
x + 135 y -1 = 0
13
` (normal form x cos φ+ y sinφ – p = 0 )
We can see from the normal form that p = 1, cos φ= ‐
12/13 and sin φ = 5/13. Since cos is negative and sin is positive, the normal lies in the second quadrant. Tan φ= ‐ 5/12
Take the Arctan and we get φ= arctan ‐5/12= ‐22.6 degrees. Since it is in the second quadrant (180‐22.6)= 157.4
Thus the polar form of 12x-5y+13=0 is 1=r cos (θ‐157.4)
`
` On
our previous equation 12 x – 5y +
13 = 0, we can determine the slope
and the y intercept.
` Placing it in slope intercept form, we
have
y = 12/5 x + 13/5
So the slope is 12/5 and the y intercept
is 13/5
Now let’s convert back. Write 5 = r cos (θ‐45) in rectangular form
We need to remember the angle sum and difference identities for
cosine and sine from lesson 7-3.
`
m
Cos (α±β) = cos α cosβ m sinα sinβ (note the )
Sin (α±β) = sinα cosβ ± cosα sinβ
So this problem we will use the difference identity for cos
5= r (cos θ cos 45 + sin θ sin 45) 5 = r ( cos θ + sin θ) 2
2
5=
10 =
2
2
r cos θ + r sin θ
2
2
2
x +
2
y
2
2
2
2
5 = x + y 2
2
2
x+
2
y -10 = 0
`
π
Graph the equation 2 = r cos (θ + ). Using 2
your calculator, set up a table of values.
θ
0
90
180
225
270
315
330
360
r
UD
-2
UD
2.83
2
2.83
4
UD
`Section
9-4
`Pp. 578-579
`#13,17,19,23,25,38
`