Missed Blog

I was sitting in Alex's dorm cross-stitching a picture of a hummingbird when I realized I didn't blog last week. I guess State burnt me out and I unconsciously decided not to think of math until I had to. Add that to my inability to remember to do things I have to do... Well anyway, here it is anyway if you would be so kind to give me half credit of some sort <3 :P

Sorry! >.<

Anyway, angles of inclination.
For lines, you will use the formula m = tan(α), where m is the slope of the line.
For conics, you will use tan 2α = (B)/(A-C) where Ax^2+Bxy+Cy^2+...blah blah blah

Your angle of inclination will be α.

For those who forget or never knew, a slope of a line in the form of ax+by = c, is (-a/b).
You could also solve the equation for y.
y = mx+b, slope is m.

You can use a handy dandy trick that works where A=C for conics. The angle of inclination will be π/4, and 5π/4.

Ex.1

Find the inclination of the line 3x+5y = 8.

• m = tanα because it is a line.

• m = -a/b, so m = -3/5.

• Plugging in, -3/5 = tanα.

• To solve, you have to take the inverse of both sides.
  tan^-1(-3/5) = α

• tan^-1(3/5) = 30.964, tan is negative in quadrants II and IV.

• tan^-1(-3/5) = 149.036 and 329.036

• α = 149°2'9'' and 329°2'9''

Ex.2
Identify the graph of x^2-xy+2y^2=2 and find the angle of inclination.

• A = 1, B = -1, C = 2

• B^2-4(A)(C) = -7

• It's negative and A does not equal C, so it's an ellipse.

• You will use tan2α = (B)/(A-C)

• tan2α = -1/-1 = 1

• Ignore the 2 in front of α for now and take the inverse.

• tan^-1(1) = 45°

• tan is positive in quadrants I and III.

• tan^-1(1) = 45° and 225°

• Then divide by 2.

• α = 45/2° and 225/2°