I was disappointed as I sat down to write this post as I didn’t get through the unit Conic Sections this week but, after going through my notes to create the outline for this post, in retrospect I realized I got through a TON of work this week and am actually really happy with my effort. I assumed this unit would be about calculating the perimeter and area of cones which I was wrong about. The word ‘conic’ (from what I’ve learned so far) refers to circles, ellipses, parabolas and hyperbolas. I spent the week reviewing/learning their respective equations/formulas and components. I was able to understand everything that was shown to me throughout the week but I don’t have a solid grasp on much of it yet. That being said, I’m confident I’ll eventually get an intuitive understanding of how it all works with a bit more practice.

The unit opened with a description of the four conic sections:

I have a good amount of experience working with circles, some experience working with ellipses and parabolas, but almost no experiences working with hyperbolas. The first shape I worked on this week was the circle where I was introduced (maybe reintroduced?) to it’s standard equation:

The above photo is probably a bit confusing but the key points are:

• Standard Equation for Circle:
  
  • (x – (h))^2 + (y – (k))^2 = r^2
    • (h, k) are the coordinates for the center of the circle.
    • (x, y) are the coordinates for ANY point on the perimeter of the circle.
    • (r) is the radius of the circle.
      
  • This equation basically states, “the (Δx, Δy)^2 at any point on the perimeter of the circle to the centre of the circle will always be equal to the radius squared.”
    • This is the same thing as saying a^2 + b^2 = c^2 where a = Δx and b = Δy and the radius equals c.
      
  • Key thing to remember is that you can always create a right triangle from any point on the perimeter of the circle to the center of the circle and use Pythag’s Theorem to calculate the radius (unless the point is on the x- or y-axis in which case r equals solely the x our y value) which is where this equation comes from.

Next, I moved on to ellipses where I first learned about their defining features:

• Vertex
  • The furthest point from the origin of the ellipse at the end of the two LARGER radii (plural for radius in case you’re like me and didn’t know that).
• Co-vertex
  • The furthest point from the origin of the ellipse at the end of the two SHORTER radii.
• Major Radius
  • One of the two larger radii on the ellipse.
• Minor Radius
  • One of the two shorter radii on the ellipse.
• (Note: from what I’ve learned so far, ellipses can be both “landscape” or “portrait”, i.e. horizontal or vertical. I’m assuming they can be on any diagonal, as well, but I haven’t been taught that yet.)

I was then shown the formula for an ellipse:

I’m close to understanding how this formula works but don’t have it fully figured out yet. It’s essentially the same formula as a circle but r^2 is replaced with a^2 (the x-radius) and b^2 (the y-radius). You divide the right side of the equation (what would have been r^2 but is replaced with a^2 and b^2) by itself resulting in the right side equalling 1 (i.e. (a^2 + b^2) / (a^2 + b^2) = 1) and then divide the x- and y-terms by their respective radiuses squared, a^2 and b^2. (I think I may actually have this wrong but it’s something close to this.)

Next I learned about ‘focii’ on an ellipse (pronounced “foe-sea”) which is the plural for ‘focus’:

First off, focii are two points along the major radii that are equidistant from the origin of the ellipse. My description of focii in the photo above doesn’t fully explain them, however. I said that the distance between the focii and the co-vertices is equal to the length of the major radii which is true but there’s more to it in that. The sum of the distance between the focii and ANY point on the ellipse is equal to 2a, a.k.a. the major radii:

I then learned how to derive the formula to calculate the distance of each focii from the origin of the ellipse:

(Note: in the photo above, consider the triangle to be part of an ellipse that’s centered at (0, 0) where line segment a is the major radii and line segment b is the minor radii.)

• Step 1
  • Consider that line a is a major radius and line b is a minor radius of an ellipse. Knowing this we can construct a right triangle from the origin, to the vertex, to the co-vertex and state that a^2 + b^2 = c^2
• Step 2
  • We’re assuming that there’s a point somewhere along the major radius that, drawing a line from the co-vertex to that point, is equal length as the major radius. The point we’re referring to is the focus. The line created can be labelled as line segment d and therefore d = a.
• Step 3
  • We can then create a new right triangle inside the existing right triangle where d is the hypotenuse and a new line segment, f, sits on the major radius. The formula for this triangle is f^2 + b^2 = d^2 but since we know that d = a we can also state that f^2 + b^2 = a^2.
• Step 4
  • We can then use algebra to go from f^2 + b^2 = a ^2 to f^2 = a^2 – b^2.
• Step 5
  • To finish off, we can take the square root of both sides of the equation leaving us with the formula for the distance of the foci which is f = √(a^2 – b^2).

I then learned about the ‘focus’ and ‘directrix’ which are two tools (I think there’s probably a better word to use than ‘tools’ but I can’t think of one) that are used to calculate parabolas:

As I stated at the bottom of the photo of my notes, what’s most important to remember is any point on a parabola is equidistant to the focus and the directrix. This is imperative to understand in order to understand the following equation of a parabola that uses the focus and the directrix:

To reiterate what I wrote at the bottom of this photo, the equation can literally be read as:

• √(y – (k))^2) = √(x – (a))^2 + (y – (b))^2)
  
• “The distance from any point on a parabola to the directrix = the distance from that point to the focus.”

Here’s an example of how to input the coordinates of a focus and directrix into the above equation and simplify the equation:

I finished the week by practicing these types of questions where I was given two out of three of either the (x, y) coordinates of a point on a parabola, the coordinates of the focus, and/or the coordinate of the directrix and then asked to solve for the third missing value. I was impressed with myself at how easily I was able to understand and work through the algebra. I had a ‘milestone-moment’ (not sure if that’s actually a phrase) when I was working through algebra that would have stopped me in my tracks a year ago but now seemed very straightforward and simple.

I’m going to push back my goal of getting through this course, Precalculus, another week. I should be able to get through this unit, Conic Sections (1340/1700 M.P.), early-ish this coming week but still want to review the following unit, Probability and Combinatorics (800/800 M.P.), which will likely take me into the following week. I’m not all the stressed about getting through this course within a certain time period which I think is the right mentality to have. I’m disappointed that It’s taking me this long to learn calculus but I’m happy with the effort I’m putting in so I’m not going to stress about it.