# Intro to Conic Sections A conic section defines the shapes or geometric expressions that can be created through slicing a shape consisting of two cones with a plane. They're much more versatile that you think! ![[Pasted image 20230825114930.png]] When a plane passes through the vertex of the two cones, it forms a **degenerate** conic. Conics can be defined with the second degree equation: $0 = ax^2 + by^2 + cx + dy + e+f$ Note that there is one more approach to conics - finding the general rule the points that lie on it satisfy. Different conic sections will have different forms of this - let's take a look at them right now! # Parabolas It's probably common knowledge to you that a parabola has a general equation: $y = ax^2+bx+c$ Pretty standard GCSE stuff - but don't be discouraged if you've forgotten it! It's time for us to press on with the parabola - onwards, with conic sections! Some questions in the GCSE ask you to complete the square. To do this, you'll typically use the **vertex form** - written as shown: $y = a(x-h)^2 + k$ Instead of using $x$-intercepts as references, you use the tip of the parabola - its *vertex*. With these two models, the erudite among you have to wonder - is there a *universal parabola model*? The more erudite among you may have realised that that is a rhetorical question! We'll introduce a familiar concept - the **focus** - into our standardised parabolic equations: $(x-h)^2 = 4p(y-k)$ $(y-k)^2 = 4p(x-h)$ The $p$ here denoted the distance from the focus to the vertex. Once again, if the vertex with coordinates $(h,k)$ is on the origin we have a couple simplified equations we can use! $x^2 = 4py$ $y^2 = 4px$ Now it's time to dive into the intuition. This'll serve us well for the rest of the conics' chapter - so please memorise this! >[!Example]- Finding Focus - Example >Given a parabola $y = x^2 + 6x - 12$ find its' focus. > >**Solution:** >We can find the focus by first completing the square - remember, we need to find the coordinates of the vertex before we can relate it to the focus! >$x^2 - 6x + 12 = (x-3)^2 + 3$ >This gives us a coordinate for our vertex at $(3,3)$. This parabola also satisfies the equation: >$(y-3)^2 = 4p(x-3)$ >We've done it! >[!Abstract]- The Latus Rectum of a Parabola # Circles A circle is a closed shape which can be defined with a constant radius $r$ and revolves around an arbitrary central point. The formula describing it is: >[!INFO] General Circle Formula >$(x-h)^2 + (y-k)^2 = r^2$ Where: - $h$ is the x-coordinate of the midpoint of the circle - $k$ is the y-coordinate of the midpoint of the circle - $r$ is the resultant radius of the circle Simple enough! There's a geometric explanation for the circle if you need some extra, of course! %%diagram + geometric explanation%% # Ellipses Ellipses, or ovals, are the constituent shapes of orbits. They're basically circles with different heights and widths - or semi-major and semi-minor axes, and with a new parameter, the eccentricity, $e$. However, strictly speaking, they're the shape traced out when a line is drawn so that the sum of the distances between two points on the same line, called *foci*, are equal. The centre of the ellipse will always be at the midpoint of the line between two foci, so if we had two foci at (1,3) and (3,5), the midpoint would be at (2,4). ![[Pasted image 20230916222333.png]] There are actually two formulae for an ellipse. Given a horizontal semi-major axis: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = r^2$ Given a vertical semi-major axis: $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = r^2$ This means that an ellipse with a centre at (0,0) will have an equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = r^2$ Here's more on using the midpoint to prove identities - if you're like me and are a physics enthusiast, you might see some parallels down the line! >[!Success]- General Ellipse Formula Proof >We introduce a new variable, $c$, that is the distance from the foci to the midpoint of the circle. This lets us find the length of the ellipse, given by two times the semi major axis $a$: >$(a+c) + (a-c) = 2a$ >Time to use our definition of the ellipse - at any point (h,k) on the ellipse, this must mean that the distance between the point and the two foci must be 2a as shown: >$\sqrt{ [x - (h - c)^2] + (y-k)^2 } + \sqrt{ [x - (h + c)^2] + (y-k)^2 } = 2a$ >Expanding this reduces it to: >$(a^2+c^2)(x-h)^2 + a^2(y-k)^2 = a^2(a^2-c^2)$ >The lengths $a^2 - c^2$ is equal to the square of the semi-minor axis $b^2$ as shown in the diagram, where a right triangle is created. Therefore, substitutions can be made: >$b^2(x-h)^2 + a^2(y-k)^2 = a^2b^2$ >$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ >[!Abstract]- More on Midpoints (TBA) > >![[Pasted image 20230917205749.png]] Finding the eccentricity is a matter of plugging in the formula: $e = \frac{c}{a}$ Where $c$ is the distance to the foci of the ellipse and $a$ is the ellipses' semi-major axis. As $c = \sqrt{a^2 - b^2}$ this also means we can get another form for the eccentricity equation: $e = \frac{\sqrt{a^2 - b^2}}{a}$ $e^2 = \frac{{a^2-b^2}}{a^2}$ $e = \sqrt{ 1 - \frac{b^2}{a^2} }$ This isn't the best way to do this - it's much easier finding the foci of an ellipse, but it's great as a shortcut regardless! >[!Danger]- More on Eccentricity (move to polar form ellipses) >Let's take an ellipse with arbitrary semi-major and semi-minor axes! Remember, these are given the variables $a$ and $b$ respectively. One last variable we can define is the >Remember that: >$c^2 = a^2-b^2$ >Let's take a particle at a certain arbitrary distance $r$ away from the midpoint of the ellipse. This is going to serve as our waypoint for the time being - we won't have to assign an actual value to it! Since we can make a right-angled triangle (as proven in the "more on midpoints") section, we can get equations for the $x$ and $y$ equations of this point $r$: >$x = r\cos \theta$ >$y = r\sin \theta$ >Another general formula of an ellipse is $\frac{(x+c)^2}{a^2} + \frac{y^2}{b^2} = 1$. Plug in our values for the $x$ and $y$ coordinates: >$\frac{(r^2\cos^2 \theta) + 2r\cos \theta + c^2}{a^2} + \frac{r^2\sin^2 \theta}{b^2} = 1$ >Cross multiply over... >$\frac{(b^2r^2\cos^2 \theta) + 2rb^2\cos \theta + b^2c^2}{a^2b^2} + \frac{a^2r^2\sin^2 \theta}{a^2b^2} = a^2b^2$ >$\frac{(b^2r^2\cos^2 \theta) + 2rb^2\cos \theta + b^2c^2 + a^2r^2\sin^2 \theta}{a^2b^2}$ >[[Trig Identities - Derivations DLC Ver. (Maths)#Trig Identities Basic|Use the trig identity]] to filter $\sin^2 \theta$ and $\cos^\theta$ out! >$\frac{b^2r^2 + a^2r^2 + 2rb^2\cos \theta + b^2c^2}{a^2b^2}$ >[!Example]- Eccentric Eccentricity Example Problem >TBA >[!Example]- Finding a period from the Eccentricity >How ## Further Ellipses - Even More Word Salad! I'll take that you've followed through with this... somehow. Admittedly, the PreCalc book I'm using is just a jumble of [[Reading Research Papers#Learning Latex Lost Me My Lover|LaTeX]] so it's been a little harder to get myself to do anything related to maths. Let's dive into some terms we missed! I'll list them out here for now, diagram below for reference. **Terms:** - *Latus Rectum* - not to be confused with a digestive system organ, it refers to a line perpendicular to the semi-major axis that intersects through a focus. Another way to find the eccentricity! - *Conjugate Axis* - another name for the semi-minor axis - *Transverse Axis* - another name for the semi-major axis - *Vertex* - where the axis of the ellipse intersects with the line of symmetry of the ellipse - *Directrix* - line drawn outside of the ellipse perpendicular to the semi-major axis %%diagram!!! !!!!!!! !!!!!!!!!!!%% Mind you, the Latus Rectum stretches across the ellipse perpendicular to the semi-major axis. To find it, let's employ this equation: $L = \frac{2b^2}{a}$ Where $b$ and $a$ denote the semi-minor and major axes. >[!Abstract]- Deriving the Equation >![[Pasted image 20231030150704.png]] > > # Hyperbolas Given a hyperbola we can generalise it's formula into: $\frac{(x-h)^2}{b^2} - \frac{(y-k)^2}{a^2} = r^2$ It's just an ellipse with a minus sign... or so you think! Let's take a closer look at one of these mysterious shapes... ![[Pasted image 20231103120834.png]] ...huh, it's completely different. While that might not be a very accurate representation of the graph as a whole, it's shape is pretty unique - something that # Polar Conic Sections # Case Study - Proving the Surface Area of a Sphere Yep, seems like you can even do this too. How delighting, the US syllabus overtakes ours for once! # Example Questions