Conic Sections 1 Cheat Sheet Edexcel FP1

𝑐𝑐 2
This chapter aims to build upon the graph sketching skills you developed in year 1. Similarly, the gradient of a rectangular hyperbola can be found by rearranging into the form 𝑦𝑦 =
𝑥𝑥
As described by the name, conic sections are planes produced by Example 2: Find the co-ordinates of the focus and the equation for the directrix of a parabola with and differentiating. If you are sitting the AS paper you do not need to know how to differentiate
slicing two cones placed point to point- more mathematically, the equation 𝑦𝑦 2 = √80𝑥𝑥. implicitly or parametrically, you will be given the result in the exam. From these results, you should
intersection of cones with planes are the graphs that you will see in be able to find a normal or tangent to a parabola or rectangular hyperbola.
this chapter. Conic sections are used in modelling: planets travel in
ellipses, the mirrors in solar power stations are parabolas and a Put the equation into the form 𝑦𝑦 2 = 4𝑎𝑎𝑎𝑎. 𝑦𝑦 2 = 4√5𝑥𝑥, so, 𝑎𝑎 = √5 Example 4: Find the normal to the parabola 𝑦𝑦 2 = 16𝑥𝑥 that passes through the point P(2,4√2).
variety of conic sections are used in engineering. In the set Define the focus and directrix using the forms Focus: (𝑎𝑎, 0) ⇒ (√5, 0) Find the gradient of the parabola (in an 𝑑𝑑𝑑𝑑 2𝑎𝑎 8
of diagrams to the right, diagram 1 shows how a circle is obtained above. Directrix: 𝑥𝑥 + 𝑎𝑎 = 0 ⇒ 𝑥𝑥 + √5 = 0 exam you will have to derive the gradient = =
from a conic section, 2 shows an ellipse, 3 shows a parabola and 4 𝑑𝑑𝑑𝑑 𝑦𝑦 𝑦𝑦
as above)
shows a hyperbola. You also need to be able to find the equation of a parabola from the focus and directrix: Gradient=
8
= √2
4√2
Find the gradient at the point P, and The gradient of the normal at a point is the
Parametric equations. Example 3: Find the equation of the parabola with focus (3,0) and directrix 𝑥𝑥 = −3
determine the gradient of the normal. negative reciprocal of the gradient at that point, so
Unlike the Cartesian form of equations that you will have seen throughout Put the equation for the directrix in the form 1
𝑥𝑥 = −3 ⇒ 𝑥𝑥 + 3 = 0 the gradient of the normal is − .
school, parametric equations are given in terms of an independent variable, 𝑥𝑥 + 𝑎𝑎 = 0. √2
called a parameter- most of the time the parameter will be denoted ‘t’, and the Define a. Focus: (3,0), directrix: 𝑥𝑥 + 3 = 0, so 𝑎𝑎 = 3 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐
equations are given in the form 𝑥𝑥 = 𝑝𝑝(𝑡𝑡), 𝑦𝑦 = 𝑞𝑞(𝑡𝑡). By substituting in specific Define the parabola- remember it is of the form −1
𝑦𝑦 2 = 12𝑥𝑥 4√2 = (2) + 𝑐𝑐
values of t you can find the co-ordinates of a point on the curve. To convert 𝑦𝑦 2 = 4𝑎𝑎𝑎𝑎. √2
parametric equations to Cartesian, you must eliminate the parameter between the equations- this 2
Use 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐 or 𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚𝑁𝑁 (𝑥𝑥 − 𝑥𝑥1 ) 4√2 + = 𝑐𝑐
can be often done by normal substitution, but keep an eye out for different trigonometric identities Questions involving parabolas will not just require you to find the focus and directrix, you may also be to find the equation of the normal. √2
that can eliminate the parameter: required to use skills you already know to find intersection points, lengths and perpendicular bisectors 𝑐𝑐 = 5√2
Equation of the normal is
Example 1: A Curve has parametric equations 𝑥𝑥 = 𝑟𝑟sin(𝑡𝑡) and 𝑦𝑦 = 𝑟𝑟cos(𝑡𝑡), 𝑡𝑡 ∈ ℝ and r is a positive Rectangular hyperbolas −1
𝑦𝑦 = 𝑥𝑥 + 5√2
constant. Find the Cartesian equation of the curve. Hyperbolas are obtained by slicing through both of the cones that are placed vertex to vertex. A √2
rectangular hyperbola is a specific type of hyperbola with asymptotes that meet at right angles. The
Notice that by squaring each equation, we can curve to the right is a rectangular hyperbola. Again, you must be able to
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use the trigonometric identity sin2 𝜃𝜃 + cos2 𝜃𝜃 = 𝑥𝑥 2 = 𝑟𝑟 2 sin2 𝑡𝑡, 𝑦𝑦 2 = 𝑟𝑟 2 cos2 𝑡𝑡 identify and work with both the Cartesian and parametric forms. Example 5: The point 𝑃𝑃 �4𝑡𝑡, � , 𝑡𝑡 ≠ 0 lies on the rectangular hyperbola 𝐻𝐻 with equation 𝑥𝑥𝑥𝑥 = 16.
𝑡𝑡
1. • A rectangular hyperbola has Cartesian equation 𝑥𝑥𝑥𝑥 = 𝑐𝑐 2 , where Find a tangent to 𝐻𝐻 at P.
Add the equations together to put in the form 𝑥𝑥 2 + 𝑦𝑦 2 = 𝑟𝑟 2 (sin2 𝑡𝑡 + cos2 𝑡𝑡) c is a positive constant
𝑥𝑥𝑥𝑥 = 16
of the identity we’re using. 𝑥𝑥 2 + 𝑦𝑦 2 = 𝑟𝑟 2 • A rectangular hyperbola has parametric equations Find the gradient of the hyperbola. 16 𝑑𝑑𝑑𝑑 16
𝑐𝑐 𝑦𝑦 = , =− 2
We now have an equation that involves 𝑥𝑥 = 𝑐𝑐𝑐𝑐, 𝑦𝑦 = , 𝑡𝑡 ∈ ℝ, 𝑡𝑡 ≠ 0. 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑥𝑥
𝑡𝑡
𝑥𝑥 and y only. r is a constant- notice that the • The asymptotes are the x-axis (𝑥𝑥 = 0) and the y-axis (𝑦𝑦 = 0), and Find the gradient of the hyperbola The gradient at P is
equation we have found is the equation of a 𝑐𝑐
a generic point can be defined as (𝑥𝑥, 𝑦𝑦) or (𝑐𝑐𝑐𝑐, ) at 𝑃𝑃 by substituting in the x-value −16 −1
circle centred at (0,0) with radius r. 𝑡𝑡 = 2
Basic exam questions with rectangular hyperbolas in will often be about intersecting of the co-ordinate given. 16𝑡𝑡 2 𝑡𝑡
of lines, bisectors and finding lengths or midpoints, and therefore will only require 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐
You should be able to sketch a curve given its parametric equations- most of the time the easiest way
skills you already know. 4 −1
to do this will be to convert to Cartesian form and then sketch, but if you are really struggling = 2 (4𝑡𝑡) + 𝑐𝑐
remember you can substitute values of t in and plot some points. Use 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑐𝑐 or 𝑦𝑦 − 𝑦𝑦1 = 𝑡𝑡 𝑡𝑡
Tangents and normals 4 4
𝑚𝑚𝑁𝑁 (𝑥𝑥 − 𝑥𝑥1 ) to find the equation of 𝑐𝑐 = +
As you will have seen previously, being able to find the gradient of a curve is essential for being able 𝑡𝑡 𝑡𝑡
Parabolas the tangent. −1 8
to find its tangent or normal. You can use parametric differentiation or implicit differentiation, which 𝑦𝑦 = 2 𝑥𝑥 +
You have seen parabolas throughout mathematics, but you need to be
were covered in pure year 2. For the general parabola 𝑦𝑦 2 = 4𝑎𝑎𝑎𝑎, the gradient is given by
𝑑𝑑𝑑𝑑 2𝑎𝑎
= , 𝑡𝑡 𝑡𝑡
able to identify and be confident working with the parametric form. 𝑑𝑑𝑑𝑑 𝑦𝑦 𝑦𝑦𝑡𝑡 2 = −𝑥𝑥 + 8𝑡𝑡
The graph to the right is the graph of 𝑦𝑦 2 = 4𝑎𝑎𝑎𝑎, where a is a positive which can be derived either parametrically or implicitly
constant. Loci
At first, this seems like a complicated way of defining a parabola, but this The set of all of the tangents gives the envelope of a curve The focus directrix property of a parabola can be used to derive its general equation.
format allows the focus-directrix properties to appear much simpler.
Example 3: Derive the gradient of a general parabola 𝑦𝑦 2 = 4𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜 𝑥𝑥 = 𝑎𝑎𝑡𝑡 2 , 𝑦𝑦 = 2𝑎𝑎𝑎𝑎 both Example 6: The curve C is the locus of points that are equidistant from the line with equation 𝑥𝑥 +
• A parabola has Cartesian equation 𝑦𝑦 2
= 4𝑎𝑎𝑎𝑎, where a is a parametrically and implicitly. 3 = 0 and the point (3,0). Prove that C is a parabola.
Positive constant. It has parametric equations 𝑑𝑑𝑑𝑑 Set up an equation using the fact that a generic
Parametric differentiation: 𝑥𝑥 = 𝑎𝑎𝑡𝑡 2 ⇒ = 2𝑎𝑎𝑎𝑎 We know from the question that the locus
𝑑𝑑𝑑𝑑 point P will be equidistant from the focus, S,
𝑥𝑥 = 𝑎𝑎𝑡𝑡 2 , 𝑦𝑦 = 2𝑎𝑎𝑎𝑎, 𝑡𝑡 ∈ ℝ. Differentiate each parametric equation with respect to t. 𝑑𝑑𝑑𝑑 satisfies 𝑺𝑺𝑺𝑺 = 𝑿𝑿𝑿𝑿.
𝑦𝑦 = 2𝑎𝑎𝑎𝑎 ⇒ = 2𝑎𝑎 and the directrix, X.
𝑑𝑑𝑑𝑑 When we talk about distances, Pythagoras’
Clearly this curve is symmetric about the x-axis, and a generic point 𝑑𝑑𝑑𝑑 1 theorem is often useful, so we square the
𝑑𝑑𝑑𝑑
can be defined as (𝑥𝑥, 𝑦𝑦)or (𝑎𝑎𝑡𝑡 2 , 2𝑎𝑎𝑎𝑎). Use similar reasoning to the chain rule to find , = values of our generic point P to get our
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2𝑎𝑎𝑎𝑎 𝑺𝑺𝑷𝑷𝟐𝟐 = 𝑿𝑿𝑷𝑷𝟐𝟐
𝑑𝑑𝑑𝑑
recalling that = 1/ .
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2𝑎𝑎 1 equation into a more useful form. We then (𝑥𝑥 − 3)2 + (𝑦𝑦 − 0)2 = (𝑥𝑥 + 3)2
As well as being viewed as graphs of functions, parabolas, and all conic sections, 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = × = =
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2𝑎𝑎𝑎𝑎 𝑡𝑡 express the distances between S and P and X
can be defined geometrically. A parabola is the locus (set of all points), whose and P - it is useful to sketch a diagram.
distance from the focus (a certain point) is equal to the distance from the directrix 𝑦𝑦 2 = 4𝑎𝑎𝑎𝑎
Implicit differentiation: 𝑑𝑑𝑑𝑑 𝑥𝑥 2 − 6𝑥𝑥 + 9 + 𝑦𝑦 2 = 𝑥𝑥 2 + 6𝑥𝑥 + 9
(a certain line). For a parabola with cartesian equation 𝑦𝑦 2 = 4𝑎𝑎𝑎𝑎: 2𝑦𝑦 = 4𝑎𝑎 Expand and simplify our expression.
𝑑𝑑𝑑𝑑 𝑦𝑦 2 = 12𝑥𝑥
The curve C has an equation of the form 𝑦𝑦 2 =
• The focus, denoted S, has co-ordinates (𝑎𝑎, 0) 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 2𝑎𝑎 State what you have found. 4𝑎𝑎𝑎𝑎, in our case 𝑎𝑎 = 3, so is therefore a
Rearrange to find , notice that by substituting in t, the
• The directrix has equation 𝑥𝑥 + 𝑎𝑎 = 0 𝑑𝑑𝑑𝑑 = parabola.
two gradients are the same, as expected. 𝑑𝑑𝑑𝑑 𝑦𝑦
• The vertex is at the origin, (0,0).

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