1. Further Pure 1 Chapter 2 :: Conics 1
Last modified: 6th September 2019

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3. Conic Sections Circle Ellipse Parabola Hyperbola
There are a number of ‘families’ of curves that can be obtained when considering the intersection of a plane with a (non-solid) ‘double cone’. For this reason, such curves are known as conics.
Circle
First Seen: GCSE/Pure Year 1
Obtained when the plane is parallel to the base of the cones.
Ellipse
First Seen: Chapter 3
The plane is less steep than the surface of the cone.
Fro Note: Conics are lines. The shaded area is not included.
Parabola
First Seen: GCSE/Pure Year 1
Plane is parallel to the surface of the cone.
Different types of curves are used in different real-life settings. e.g. The hanging cable of a suspension bridge is hyperbolic if hanging freely, and parabolic if it is attached by supporting cables to the road below.
Hyperbola
First Seen: Chapter 3
The plane is steeper than the surface of the cone, so intersects both cones.

4. RECAP :: Parametric vs Cartesian Equations
A Cartesian equation is one which says how the 𝑥, 𝑦 (𝑧 etc.) values are related for any point on the line.
Examples: 𝑥+2𝑦=1
𝑥 2 −3 𝑦 2 =2
A parametric equation is one where each of 𝑥, 𝑦 (𝑧 etc.) are expressed in terms of an independent variable (known as the parameter).
Examples: 𝑥= tan 2 𝜃
𝑦= sin 𝜃

5. Example
(This has been used as a Computer Science interview question at both Oxford and Cambridge) 𝑥
A ladder of length 2𝑙 is initially vertical up against a wall. The ladder gradually slides down to the floor.
Determine the equation of the trajectory of the midpoint of the ladder. 𝑙
Cartesian Equation:
(Hint: Pythagoras)
(𝑥,𝑦) ?
If we draw a line down from (𝑥,𝑦), we get a right-angled triangle.
𝒙 𝟐 + 𝒚 𝟐 = 𝒍 𝟐
i.e. The trajectory is a circle with radius 𝑙. 𝑙  𝑦
Parametric Equation:
(Based on a parameter  we could introduce for the angular inclination of the ladder)
Cartesian Variables: 𝒙, 𝒚
Parametric Variables: 𝜃
Constants: 𝒍 ? ? ? ?
By simple trigonometry:
𝑥= cos 𝜃 , 𝑦= sin 𝜃

6. Sketching Parametric Curves
Sketch the curve with equation 𝑥=𝑎 𝑡 2 , 𝑦=2𝑎𝑡, 𝑡∈ℝ, where 𝑎 is a positive constant.
Let’s just try a few values for the parameter and see what coordinates we get... 𝑡 −3 −2 −1 1 2 3 𝑥 9𝑎 4𝑎 𝑎 𝑦
−6𝑎
−4𝑎
−2𝑎 2𝑎 6𝑎 ?
? Sketch 6𝑎 4𝑎 2𝑎 𝑂
-2𝑎
-4𝑎
−6𝑎
2𝑎 𝑎 𝑎 𝑎

7. RECAP :: Converting Parametric  Cartesian
We can use either substitution or elimination to turn parametric equations into a Cartesian one.
Find the Cartesian equation for the parametric equations 𝑥=𝑎 𝑡 2 , 𝑦=2𝑎𝑡, where 𝑎 is a positive constant. ?
𝒕= 𝒚 𝟐𝒂 → 𝒙=𝒂 𝒚 𝟐𝒂 𝟐 → 𝒚 𝟐 =𝟒𝒂𝒙
Find the Cartesian equation for the parametric equations 𝑥=𝑐𝑡, 𝑦= 𝑐 𝑡 , where 𝑐 is a positive constant. ?
𝒕= 𝒙 𝒄 → 𝒚= 𝒄 𝒙/𝒄 = 𝒄 𝟐 𝒙
Alternatively we could have multiplied the two equations to obtain 𝒙𝒚= 𝒄 𝟐

8. Exercise 2A Pearson Further Pure 1 Pages 34-35
(Classes in a hurry may wish to skip this exercise)

9. RECAP :: Parabola 𝒚=𝒂 𝒙 𝟐 +𝒃𝒙+𝒄 𝒚=𝒂 𝒙 𝟐 𝒚𝟐 = 𝒂𝒙
You may already be familiar that the name of the line governed by a quadratic equation is known as a parabola. For any vertically aligned parabola (where line of symmetry is vertical):
𝒚=𝒂 𝒙 𝟐 +𝒃𝒙+𝒄
If we consider just those just centred at the origin, we know its equation will be of the form:
𝒚=𝒂 𝒙 𝟐
However, in this chapter we’ll only be considering parabolas which are horizontally aligned, by just swapping 𝑥 and 𝑦. We’ll also only consider those where the constant 𝑎 is positive:
𝒚𝟐 = 𝒂𝒙 𝑥 𝑦 𝑦 𝑥 𝑥 𝑦 𝑥

10. Definition of parabola?
! No need to write this down
A locus of points is a set of points satisfying a certain condition. ?
So a parabola (by definition) is a locus of points equidistant from a line (directrix) and point (focus).
Loci involving:
Thing A
Thing B
Interpretation
Resulting Locus of points
Point -
A given distance from point A A
Reveal
Line -
A given distance from line A A
Reveal
Perpendicular bisector
Point
Point
Equidistant from 2 points.
Reveal A B
In the context of a parabola, we call this line the directrix.
Line
Line
Equidistant from 2 lines
Angle bisector A
Reveal
...and this point the focus of the parabola. B
Point
Line
Equidistant from point A and line B A
Parabola
Reveal B

11. Parabola 𝒙=𝒂 𝒕 𝟐 , 𝒚=𝟐𝒂𝒕, 𝒕∈ℝ 𝒚 𝟐 =𝟒𝒂𝒙 ! Write all this down
A horizontally-aligned parabola centred at the origin has equations:
(for some positive constant 𝑎)
Parametric:
𝒙=𝒂 𝒕 𝟐 , 𝒚=𝟐𝒂𝒕, 𝒕∈ℝ
Cartesian:
𝒚 𝟐 =𝟒𝒂𝒙
A general point 𝑃 on this curve has coordinates 𝑃 𝑥,𝑦 or 𝑃 𝑎 𝑡 2 , 2𝑎𝑡 𝑦
A parabola is a locus of points such that the distance from any point to the focus is the same as the distance to the directrix.
𝑃 𝑥,𝑦
𝑥 = −𝑎
Focus: 𝒂,𝟎
Directrix: Line 𝑥=−𝑎
Vertex: 𝟎,𝟎 𝑥 −𝑎 𝑎
FOCUS
VERTEX
AXIS OF SYMMETRY
DIRECTRIX

12. Equations of Parabolas
Find an equation of the parabola with focus (7,0) and directrix 𝑥+7=0 ?
𝒂=𝟕 → 𝒚 𝟐 =𝟐𝟖𝒙 Q
... and with focus ,0 and directrix 𝑥=− ?
𝒚 𝟐 = 𝟑 𝒙 Q
Find the coordinates of the focus and an equation for the directrix of a parabola with equation:
𝑦 2 =24𝑥
𝑦 2 = 32 𝑥
Focus: 𝟔,𝟎 Directrix: 𝒙=−𝟔
Focus: 𝟐 ,𝟎 Directrix: 𝒙=− 𝟐 ? ?
Quickfire Questions:
Equation:
𝑦 2 =16𝑥
𝑦 2 =100𝑥
𝑦 2 =24𝑥
𝑥 2 =12𝑦
𝑦= 𝑥 2
Focus:
(4, 0)
(25,0)
(6,0)
(0,3)
(0, 0.25)
Directrix:
𝑥=−4
𝑥=−25
𝑥=−6
𝑦=−3
𝑦=−0.25 ? ? ? ? ? ?
You wouldn’t be asked this in an exam. ? ? ? ?

13. Exercise 2B Pearson Further Pure 1 Page 36
(Again, classes in a hurry may wish to skip this very short exercise)

14. Coordinate Geometry involving Parabolas
Question: A point 𝑃(8, −8) lies on the parabola C with equation 𝑦 2 =8𝑥. The point 𝑆 is the focus of the parabola. The line l passes through S and P.
Find the coordinates of 𝑆. 𝟐,𝟎 , as we just quarter the 8.
Find a equation for 𝑙, giving your answers in the form 𝑎𝑥+𝑏𝑦+𝑐=0, where 𝑎, 𝑏, 𝑐 are integers. 𝒎=− 𝟖 𝟔 =− 𝟒 𝟑 Using point 𝑺: 𝒚−𝟎=− 𝟒 𝟑 𝒙−𝟐 → 𝟒𝒙+𝟑𝒚−𝟖=𝟎
The line 𝑙 meets the parabola 𝐶 again at the point 𝑄. The point 𝑀 is the mid-point of 𝑃𝑄. Find the coordinates of 𝑄. 𝒍: 𝟒𝒙+𝟑𝒚−𝟖=𝟎 C: 𝒚 𝟐 =𝟖𝒙 Solving simultaneously gives us (𝟎.𝟓, 𝟐)
Find the coordinates of 𝑀. 𝟏𝟕 𝟒 , −𝟑
Draw a sketch showing parabola 𝐶, the line 𝑙 and the points 𝑃, 𝑄, 𝑆 and 𝑀. ? ? ? ? 𝑦
C: 𝑦 2 =8𝑥
Q(0.5, 2) ?
S(2, 0) 𝑥
M(17/4, -3)
L: 4𝑥+3𝑦−8=0
P(8, -8)

15. Further Example
The parabola 𝐶 has general point 𝑎 𝑡 2 ,2𝑎𝑡 . The line 𝑥=𝑘 intersects 𝐶 at the points 𝑃 and 𝑄. Find, in terms of 𝑎 and 𝑘, the length of the chord 𝑃𝑄. ?
𝑥=𝑎 𝑡 → 𝑘=𝑎 𝑡 2 𝑡=± 𝑘 𝑎
𝑦=2𝑎𝑡=±2𝑎 𝑘 𝑎 =±2 𝑎𝑘
Coordinates are:
𝑃 𝑘,2 𝑎𝑘 and 𝑄 𝑘,−2 𝑎𝑘
Length of chord 𝑃𝑄 is 4 𝑎𝑘

16. Test Your Understanding
Edexcel FP1(Old) Jan 2009 Q8 ? ? ? ?

17. Exercise 2C
Pearson Further Pure 1
Pages 39-42

18. Rectangular Hyperbolas
As mentioned at the start of the chapter, if the intersecting plane is steeper than the surface of the cone, we intersect both cones, producing a hyperbola.
We will fully explore these in Chapter 3, but in this chapter we will look at a special case, rectangular hyperbolas, where the plane is vertical.

19. Chapter 3 preview
Source: Wikipedia
We know that the equation of a circle with unit radius is:
𝒙 𝟐 + 𝒚 𝟐 =𝟏
The corresponding hyperbola would be:
𝒙 𝟐 − 𝒚 𝟐 =𝟏
and would look like the red curves on the left.
A hyperbola has two focal points (F1 and F2) rather than one, and two directrices D1 and D2.
Unlike parabolas, where the distance to the directrix and focus was equal, there’s now a (constant) factor difference, denoted by 𝑒 (known as the ‘eccentricity’).

20. Rectangular Hyperbolas
! A rectangular hyperbola is a hyperbola whose asymptotes are perpendicular.
𝑥 2 4 − 𝑦 2 9 =1
Example: 𝑥 2 − 𝑦 2 =1
The asymptotes of a hyperbola are not necessarily perpendicular
𝑦= 3 2 𝑥
𝑦=− 3 2 𝑥

21. Rectangular Hyperbolas
In this chapter, you only need to know about rectangular hyperbola whose asymptotes are vertical and horizontal.
You previously identified this type of equation at GCSE as a:
Recriprocal equation! ?
Source: WolframAlpha.com
“Reciprocal graph” is not a formal mathematical name for the line represented by equation 𝒚= 𝒌 𝒙 .
Look what happens if you type 𝑦= 1 𝑥 in WolframAlpha!:

22. Rectangular Hyperbola
! Write all this down
A rectangular hyperbola with asymptotes 𝑥=0 and 𝑦=0, has the equations:
Parametric:
𝒙=𝒄𝒕, 𝒚= 𝒄 𝒕 , 𝒕∈ℝ, 𝒕≠0
Cartesian:
𝒚= 𝒄 𝟐 𝒙 or 𝒙𝒚= 𝒄 𝟐
where 𝑐 is a positive constant.
A general point P on this curve has coordinates 𝑃 𝑥,𝑦 or 𝑃 𝑐𝑡, 𝑐 𝑡 𝑦
(Note: You do not [yet] need to know how to find the vertices, foci or directrices) 𝑥

23. Example
[Textbook] The rectangular hyperbola 𝐻 has Cartesian equation 𝑥𝑦=64. The line 𝑙 with equation 𝑥+2𝑦−36=0 intersects the curve at the points 𝑃 and 𝑄.
Find the coordinates of 𝑃 and 𝑄.
Find the equation of the perpendicular bisector of 𝑃𝑄 in the form 𝑦=𝑚𝑥+𝑐. ?
𝒙+𝟐𝒚−𝟑𝟔=𝟎 → 𝒙=−𝟐𝒚+𝟑𝟔 −𝟐𝒚+𝟑𝟔 𝒚=𝟔𝟒 −𝟐 𝒚 𝟐 +𝟑𝟔𝒚−𝟔𝟒=𝟎 𝒚 𝟐 −𝟏𝟖𝒚+𝟑𝟐=𝟎 𝒚−𝟏𝟔 𝒚−𝟐 =𝟎 𝒚=𝟐 ⇒ 𝒙=𝟑𝟐 ⇒ 𝑷 𝟑𝟐,𝟐 𝒚=𝟏𝟔 ⇒ 𝒙=𝟒 ⇒ 𝑸(𝟒,𝟏𝟔)
Midpoint of 𝑷𝑸 is (𝟏𝟖,𝟗)
Gradient of 𝑷𝑸 is − 𝟏 𝟐
Gradient of perpendicular bisector is 𝟐
𝒚−𝟗=𝟐 𝒙−𝟏𝟖 𝒚=𝟐𝒙−𝟐𝟕
There’s nothing new here. Just use the usual Pure Year 1 theory, e.g. simultaneous equations for points of intersection.

24. Exercise 2D
Pearson Further Pure 1
Page 44

25. Equations of Tangents and Normals
The point 𝑃, where 𝑥=2, lies on the rectangular hyperbola 𝐻 with equation 𝑥𝑦=8. Find:
The equation of the tangent 𝑇.
The equation of the normal 𝑁
to 𝐻 at the point 𝑃, giving your answer in the form 𝑎𝑥+𝑏𝑦+𝑐=0. ?
𝒚=𝟖 𝒙 −𝟏
𝒅𝒚 𝒅𝒙 =− 𝟖 𝒙 𝟐
When 𝒙=𝟐 ⇒ 𝒚=𝟒, 𝒎 𝑻 =−𝟐, 𝒎 𝑵 = 𝟏 𝟐
Equation of tangent:
𝒚−𝟒=−𝟐 𝒙−𝟐 𝟐𝒙+𝒚−𝟖=𝟎
Equation of normal:
𝒚−𝟒= 𝟏 𝟐 𝒙−𝟐 𝒙−𝟐𝒚+𝟔=𝟎
Make 𝑦 the subject so we can find 𝑑𝑦 𝑑𝑥
Use 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1 as per usual to find equation of tangent.

26. Equations of Tangents and Normals
The distinct points A and B, where 𝑥=3 lie on the parabola C with equation 𝑦 2 =27𝑥.
The line 𝑙 1 is the tangent to C at A and the line 𝑙 2 is the tangent to C at B. Given that at A, 𝑦>0,
Find the coordinates of A and B.
Draw a sketch showing the parabola C. Indicate A, B, 𝑙 1 and 𝑙 2 .
Find equations for 𝑙 1 and 𝑙 2 , giving your answer in the form 𝑎𝑥+𝑏𝑦+𝑐=0.
y2 = 81, so y = 9 ?
c) 𝑦=± 3 𝑥 𝑑𝑦 𝑑𝑥 =± 𝑥 𝑚 𝑙 1 = 3 2 … 𝑙 1 :3𝑥−2𝑦+9=0 𝑙 2 :3𝑥+2𝑦+9=0 ? ? l1
B(3,9)
B(3,-9) l2

27. Using implicit differentiation
Those who have covered Pure Year 2 content will have encountered implicit differentiation, i.e. when we differentiate an equation involving an entangled mix of 𝑥 and 𝑦, rather than 𝑦 in terms of 𝑥. ?
A bit of cheeky product rule.
𝑑 𝑑𝑥 𝑦 2 =𝟐𝒚 𝒅𝒚 𝒅𝒙 𝑑 𝑑𝑥 𝑥𝑦 =𝒙 𝒅𝒚 𝒅𝒙 +𝒚 ?
Recap: Q
The point 𝑃 with coordinates 75,30 lies on the parabola 𝐶 with equation 𝑦 2 =12𝑥.
Find the equation of the tangent to 𝐶 at 𝑃, giving your answer in the form 𝑦=𝑚𝑥+𝑐 ?
𝒚 𝟐 =𝟏𝟐𝒙 𝟐𝒚 𝒅𝒚 𝒅𝒙 =𝟏𝟐 → 𝒅𝒚 𝒅𝒙 = 𝟔 𝒚
When 𝒚=𝟑𝟎, 𝒎= 𝟔 𝟑𝟎 = 𝟏 𝟓
𝒚−𝟑𝟎= 𝟏 𝟓 𝒙−𝟕𝟓 𝒚= 𝟏 𝟓 𝒙+𝟕𝟓
Use implicit differentiation.

28. Exercise 2E Pearson Further Pure 1 Pages 48-49
(Note: Historically most exam questions have been based on the next section rather than this one)

29. Tangents/Normal with Parametric Differentiation
The point 𝑃(𝑎 𝑡 2 ,2𝑎𝑡) lies on the parabola 𝐶 with equation 𝑦 2 =4𝑎𝑥 where 𝑎 is a positive constant. Show that an equation of the normal to 𝐶 at 𝑃 is 𝑦+𝑡𝑥=2𝑎𝑡+𝑎 𝑡 3
Recall that 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡
We could use parametric differentiation or differentiate the Cartesian equation.
It might be confusing to have to many different letters. It’s helpful to be clear what each are and what type:
𝑥 and 𝑦 are variables represent points (𝑥,𝑦) on the line.
𝑡 is a parametric variable allowing us to get points (𝑥,𝑦) as 𝑡 varies. 𝑥=𝑎 𝑡 2 and 𝑦=2𝑎𝑡.
𝑎 is a constant. It is fixed for a particular parabola allowing us to scale it.
Cartesian:
𝑦 2 =4𝑎𝑥 2𝑦 𝑑𝑦 𝑑𝑥 =4𝑎 𝑑𝑦 𝑑𝑥 = 4𝑎 2𝑦 = 2𝑎 𝑦
At 𝑃 𝑎𝑡 2 ,2𝑎𝑡 : 𝑚= 2𝑎 2𝑎𝑡 = 1 𝑡
Parametric:
𝑥=𝑎 𝑡 𝑦=2𝑎𝑡 𝑑𝑥 𝑑𝑡 =2𝑎𝑡 𝑑𝑦 𝑑𝑡 =2𝑎
𝑑𝑦 𝑑𝑥 = 2𝑎 2𝑎𝑡 = 1 𝑡
? Gradient
? Gradient
? Eqn of normal
𝑚 𝑁 =−𝑡 𝑦−2𝑎𝑡=−𝑡 𝑥−𝑎 𝑡 2 𝑦−2𝑎𝑡=−𝑡𝑥+𝑎 𝑡 3 𝑦+𝑡𝑥=2𝑎𝑡+𝑎 𝑡 3
Fro Tip: Your gradient should be in terms of the parameter 𝑡. Remember that we find the gradient at a specific point, even if the point is generically expressed using parametric eqns!
In general, for a parabola 𝒚 𝟐 =𝟒𝒂𝒙 the equation of the tangent/normal at 𝒂 𝒕 𝟐 ,𝟐𝒂𝒕 is:
Tangent: 𝒕𝒚=𝒙+𝒂 𝒕 𝟐
Normal: 𝒚+𝒕𝒙=𝟐𝒂𝒕+𝒂 𝒕 𝟑
But it is not worthwhile memorising these.
Use the parametric equations as your point ( 𝑥 1 , 𝑦 1 ) i.e. 𝑎 𝑡 2 ,2𝑎𝑡

30. Problem Solving ? ? b a 𝑦 2 =20𝑥 2𝑦 𝑑𝑦 𝑑𝑥 =20 so 𝑑𝑦 𝑑𝑥 = 10 𝑦
[Textbook] The parabola 𝐶 has equation 𝑦 2 =20𝑥. The point 𝑃 5 𝑝 2 ,10𝑝 is a general point on 𝐶. The line 𝑙 is normal to 𝐶 at the point 𝑃.
Show that an equation for 𝑙 is 𝑝𝑥+𝑦=10𝑝+5 𝑝 3
The point 𝑃 lies on 𝐶. The normal to 𝐶 at 𝑃 passes through the point 30,0 as shown on the diagram. The region 𝑅 is bounded by this line, the curve 𝐶 and the 𝑥-axis.
(b) Given that 𝑃 lies in the first quadrant, show that the area of the shaded region 𝑅 is 𝐶 𝑃 𝑥 𝑂
30,0 𝑙 b ?
At 30,0 , 30𝑝=10𝑝+5 𝑝 3
5 𝑝 3 −20𝑝=0 𝑝 𝑝+2 𝑝−2 =0 𝑝=0 𝑜𝑟 −2 𝑜𝑟 2 a ?
𝑦 2 =20𝑥 2𝑦 𝑑𝑦 𝑑𝑥 =20 so 𝑑𝑦 𝑑𝑥 = 10 𝑦
At 𝑃 5 𝑝 2 ,10𝑝 , 𝑑𝑦 𝑑𝑥 = 10 10𝑝 = 1 𝑝
𝑚 𝑁 =−𝑝
𝑦−10𝑝=−𝑝 𝑥−5 𝑝 2 → … → 𝑝𝑥+𝑦=10𝑝+5 𝑝 3 𝑦
As 𝑝 varies, we have different points on the parabola and therefore different normal. Substituting each 𝑝 into the equation of 𝑙 allows us to see which value of 𝑝 we want. 𝐶 𝑃
𝑝=2
𝑝=0 𝑥 𝑂
30,0
𝑝=−2
Continued…

31. Problem Solving ? Continued… b Point 𝑃 is (5 𝑝 2 ,10𝑝) 𝑝=2 → (20,20)𝑦
[Textbook] The parabola 𝐶 has equation 𝑦 2 =20𝑥. The point 𝑃 5 𝑝 2 ,10𝑝 is a general point on 𝐶. The line 𝑙 is normal to 𝐶 at the point 𝑃.
Show that an equation for 𝑙 is 𝑝𝑥+𝑦=10𝑝+5 𝑝 3
The point 𝑃 lies on 𝐶. The normal to 𝐶 at 𝑃 passes through the point 30,0 as shown on the diagram. The region 𝑅 is bounded by this line, the curve 𝐶 and the 𝑥-axis.
(b) Given that 𝑃 lies in the first quadrant, show that the area of the shaded region 𝑅 is 𝐶 𝑃 𝑥 𝑂
30,0 𝑙
Continued… b ? 𝑦
Point 𝑃 is (5 𝑝 2 ,10𝑝)
𝑝=2 → (20,20)
Using 𝑦= 𝑥 1 2
𝑅 1 = 𝑥 𝑑𝑥 =…= 𝑅 2 = 1 2 𝑏ℎ= 1 2 ×10×20=100 𝑅= = 𝐶 𝑃
𝑅 1
𝑅 2 𝑥 𝑂 20 30 𝑙

32. Test Your Understanding
FP1 (Old Spec) Jan 2009 Q8a,b ? ?

33. Exercise 2F Pearson Further Pure 1 Pages 52-54
(Note: Given the type of exam questions on this chapter historically, it’s worth spending a considerable amount of time on this exercise)

34. RECAP of Parabola Definition
At the start of the chapter we saw that the definition of a parabola is:
a locus of points such that the distance from any point to the focus is the same as the distance to the directrix. 𝑦
𝑃 𝑥,𝑦
𝑥 = −𝑎
Focus: 𝑎,0
Directrix: Line 𝑥=−𝑎
Vertex: 0,0 𝑥 −𝑎 𝑎
FOCUS
VERTEX
AXIS OF SYMMETRY
DIRECTRIX

35. Loci Proof
Can we prove that the equation of a parabola with locus (𝑎,0) and directrix 𝑥=−𝑎 is
𝑦 2 =4𝑎𝑥?
(Hint: express algebraically the distances 𝑃𝑋 and 𝑃𝑆) 𝑦
𝑃𝑋=𝑥+𝑎 𝑃𝑆= 𝑥−𝑎 2 + 𝑦 2
𝑥−𝑎 2 + 𝑦 2 = 𝑥+𝑎 2 𝑥 2 −2𝑎𝑥+ 𝑎 2 + 𝑦 2 = 𝑥 2 +2𝑎𝑥+ 𝑎 2 𝑦 2 =4𝑎𝑥 ? 𝑋
𝑃(𝑥,𝑦) 𝑆 𝑥 𝑎
𝑥 = −𝑎
A challenge for your own time:
Can you generalise this, and find the parabola for any focus (𝑞,𝑟) and any directrix 𝑦=𝑎𝑥+𝑏?

36. Further Example
The point 𝑃 lies on a parabola with equation 𝑦 2 =4𝑎𝑥.
Show that the locus of the midpoints of 𝑂𝑃 is a parabola. ?
𝑂 0,0 𝑃 𝑎 𝑡 2 ,2𝑎𝑡 ∴ Midpoint: 𝑎 𝑡 2 , 𝑎𝑡
If 𝑥= 1 2 𝑎 𝑡 2 and 𝑦=𝑎𝑡 :
𝑦 2 = 𝑎 2 𝑡 2 =2𝑎𝑥

37. Exercise 2G
Pearson Further Pure 1
Pages 55-56