1. FP3 Chapter 2 Further Coordinate Systems
Dr J Frost
Last modified: 12th December 2016

2. Recap of Conic Sections
The axis of the parabola is parallel to the side of the cone.
All the ones you’ll see can be obtained by taking ‘slices’ of a cone (known as a conic section).
C2:
Circles
FP1:
Rectangular Hyperbolas
FP3
Ellipses
FP1:
Parabolas
FP3:
Hyperbolas

3. Overview We saw circles and ‘rectangular hyperbolas’ back in FP1.
Parabolas – FP1
Circles – FP1
Hyperbolas – FP3
𝑥=−𝑎 𝑦
𝑃(𝑥,𝑦) 𝑎 𝑋 𝑆 𝑥
Picture: Wikipedia 1
𝑥,𝑦 𝜃 1
Ellipses – FP3
In FP1 you looked at both the Cartesian and parametric form of parabolas and circles, and did some coordinate geometry in terms of tangents/normal.
In FP3, you’ll have the same types of questions, except looking at ellipses and hyperbolas instead.
𝑥,𝑦 𝑏 𝜃 𝑎

4. Standard Ellipse ? ? ? ? 𝑏 1 𝑎 𝑥,𝑦 𝑥,𝑦 Cartesian equation:𝜃 1 𝑥 𝑦
𝑥,𝑦 𝜃 𝑎 𝑏 𝑥 𝑦
Cartesian equation:
𝒙 𝟐 + 𝒚 𝟐 =𝟏
Bro Hint: We seem to be scaling 𝑥 values by 𝑎 and 𝑦 values by 𝑏 on the sketch. Think about function transformations: If we originally had 𝑓(𝑥) what would give us a stretch on the 𝑥-axis?
Cartesian equation:
𝒙 𝟐 𝒂 𝟐 + 𝒚 𝟐 𝒃 𝟐 =𝟏 ? ?
Parametric equations:
𝒙= 𝐜𝐨𝐬 𝜽 𝒚= 𝐬𝐢𝐧 𝜽
Parametric equations:
𝒙=𝒂 𝐜𝐨𝐬 𝜽 𝒚=𝒃 𝐬𝐢𝐧 𝜽 ? ?

5. Using equation of ellipse
The ellipse 𝐸 has equation 4 𝑥 2 +9 𝑦 2 =36.
a) Sketch 𝐸.
b) Write down its parametric equations.
The ellipse 𝐸 has parameter equations:
𝑥=3 cos 𝜃 𝑦=5 sin 𝜃
Determine its Cartesian equation. ? ?
𝑥 𝑦 =1
4 𝑥 𝑦 =1 𝑥 𝑦 2 4 =1
𝑎=3, 𝑏=2
You want 1 on RHS. 3 2 𝑥 𝑦 −3 −2
Sketch ?
? b
𝑥=3 cos 𝜃 𝑦=2 sin 𝜃

6. Reminder: Tangents/Normals
This is an example question from FP1: Q
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
𝑦=2 𝑎 𝑥 𝑑𝑦 𝑑𝑥 = 𝑎 𝑥
At the point 𝑃, 𝑥=𝑎 𝑡 2 , ∴𝑚= 𝑎 𝑎 𝑡 2 = 1 𝑡 ∴ 𝑚 𝑁 =−𝑡
𝑦−2𝑎𝑡=−𝑡 𝑥−𝑎 𝑡 2 … 𝑦+𝑡𝑥=2𝑎𝑡+𝑎 𝑡 3
We found the gradient function (using the original Cartesian equation).
The parametric equations of the curve were:
𝑥=𝑎 𝑡 2 𝑦=2𝑎𝑡
By fixing a 𝑡 we get a specific point. As with C1, we get a specific gradient 𝑚 by subbing in a specific value of 𝑥.
Don’t be upset by the fact that our equation involves 3 variables (𝑥, 𝑦, 𝑡). When we fix the parameter 𝑡 to a specific value, this gives a straight line equation just in terms of 𝑥 and 𝑦 (and constant 𝑎)

7. Tangents/normal of ellipses
Find the equation of the tangent to the ellipse with equation 𝑥 𝑦 2 4 =1 at the point 3 cos 𝜃 ,2 sin 𝜃
𝑑𝑦 𝑑𝑥 = 𝑑𝑦/𝑑𝜃 𝑑𝑥/𝑑𝜃 = 2 cos 𝜃 −3 s𝑖𝑛 𝜃 ?
In C3 you learnt how to get the gradient function using the parametric equations directly. This saves a lot of effort!
We could have also used implicit differentiation.
𝑦−2 sin 𝜃 = 2 cos 𝜃 −3 sin 𝜃 𝑥−3 cos 𝜃 … 3𝑦 sin 𝜃 +2𝑥 cos 𝜃 =6 ? ?
Use 𝑦− 𝑦 1 =𝑚 𝑥− 𝑥 1
Show that the equation of the normal to the ellipse with equation 𝑥 2 𝑎 𝑦 2 𝑏 2 =1 at the point 𝑃 𝑎 cos 𝜃 ,𝑏 sin 𝜃 is 𝑏𝑦 cos 𝜃 =𝑎𝑥 sin 𝜃 + 𝑏 2 − 𝑎 2 cos 𝜃 sin 𝜃
𝑑𝑦 𝑑𝑥 = 𝑏 cos 𝜃 −𝑎 s𝑖𝑛 𝜃 ∴ 𝑚 𝑁 = 𝑎 sin 𝜃 𝑏 cos 𝜃 𝑦−𝑏 sin 𝜃 = 𝑎 sin 𝜃 𝑏 cos 𝜃 𝑥−𝑎 cos 𝜃 … 𝑏𝑦 cos 𝜃 =𝑎𝑥 sin 𝜃 + 𝑏 2 − 𝑎 2 cos 𝜃 sin 𝜃 ?

8. Tangents/normal of ellipses
The point 𝑃 2, lies on the ellipse 𝐸 with parametric equations 𝑥=4 cos 𝜃 , 𝑦=3 sin 𝜃.
(a) Find the value of 𝜃 at the point 𝑃.
(b) Find the equation of the normal to the ellipse at point 𝑃.
a ?
4 cos 𝜃 =2 → 𝜃= 𝜋 3 , 5𝜋 3 , … 3 sin 𝜃 = → 𝜃= 𝜋 3 , 2𝜋 3 , …
Only 𝜃= 𝜋 3 matches both.
𝑑𝑦 𝑑𝑥 = 3 cos 𝜃 −4 sin 𝜃
At 𝑃, 𝑚 𝑁 = 4 sin 𝜋 cos 𝜋 3 = 4× × 1 2 =
𝑦− = 𝑥−2
Bro Note: For a particular 𝑥 there are multiple values of 𝜃. Thus we need to choose the one that gives the same 𝑥 AND 𝑦. 𝑥
b ?

9. Simultaneous Cartesian Equations𝑥 𝑦
If a line was a TANGENT to an ellipsis (and clearly doesn’t meet the curve again), what we expect when we solve the two equations simultaneously?
Only one point of contact, so combined equation will only have one solution. ?
Show that the condition for 𝑦=𝑚𝑥+𝑐 to be a tangent to the ellipse 𝑥 2 𝑎 𝑦 2 𝑏 2 =1 is 𝑎 2 𝑚 2 + 𝑏 2 = 𝑐 2 ?
𝑥 2 𝑎 𝑚𝑥+𝑐 2 𝑏 2 =1 … 𝑥 2 𝑏 2 + 𝑎 2 𝑚 𝑎 2 𝑚𝑐𝑥+ 𝑎 2 𝑐 2 − 𝑏 2 =0
Discriminant is 0 if one solution:
2 𝑎 2 𝑚𝑐 2 =4 𝑏 2 + 𝑎 2 𝑚 2 𝑎 2 𝑐 2 − 𝑏 2 … 𝑎 2 𝑚 2 + 𝑏 2 = 𝑐 2

10. Exercises ? ? ? ? ? ? Exercise 2A Exercise 2B
Sketch the following ellipses, and find their parametric equations.
(i) 𝑥 2 +4 𝑦 2 = (ii) 4 𝑥 2 + 𝑦 2 = (iii) 𝑥 2 +9 𝑦 2 =25
𝒙=𝟒 𝐜𝐨𝐬 𝜽 , 𝒚=𝟐 𝐬𝐢𝐧 𝜽 𝒙=𝟑 𝒄𝒐𝒔 𝜽 , 𝒚=𝟔 𝒔𝒊𝒏 𝜽 𝒙=𝟓 𝒄𝒐𝒔 𝜽 , 𝒚= 𝟓 𝟑 𝒔𝒊𝒏 𝜽 ? ? ?
Exercise 2B
All questions. ? ? ?

11. Hyperbolas Standard Hyperbola: 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1 −𝑎 𝑎
𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1
(The only difference from the equation of an ellipse is the minus!) −𝑎 𝑎
Parametric equations:
𝑥=𝑎 cosh 𝑡 , 𝑦=𝑏 sinh 𝑡
Key Properties:
Because cosh 2 𝑡 − sinh 2 𝑡 =1
But since:
sec 2 𝜃 − tan 2 𝜃 =1
Is a similar form, we alternatively get:
𝑥=𝑎 sec 𝜃 , 𝑦=𝑏 tan 𝜃
When 𝑦=0, 𝑥 2 = 𝑎 2 → 𝑥=±𝑎
As 𝑥→±∞, 𝑥 2 𝑎 2 ≈ 𝑦 2 𝑏 2
Equation of asymptotes: 𝑦=± 𝑏 𝑎 𝑥

12. Hyperbolas –Questions𝑦
9 𝑥 2 −4 𝑦 2 =36
𝟗 𝒙 𝟐 𝟑𝟔 − 𝟒 𝒚 𝟐 𝟑𝟔 =𝟏 𝒙 𝟐 𝟒 − 𝒚 𝟐 𝟗 =𝟏 𝒂=𝟐, 𝒃=𝟑 −2 2 𝑥
sketch
Cartesian Equation
Parametric Equations
Eqn of Asymptotes
Sketch
𝑥 2 −4 𝑦 2 =16
𝑥=4 cosh 𝑡 , 𝑦=2 sinh 𝑡
𝑦=± 1 2 𝑥
4 𝑥 2 −25 𝑦 2 =100
𝑥=5 cosh 𝑡 𝑦=2 sinh 𝑡
𝑦=± 2 5 𝑥
𝑥 2 8 − 𝑦 2 2 =1
𝑥=2 2 cosh 𝑡 𝑦= 2 sinh 𝑡 𝑦 ? ? ? 𝑥 −4 4 ? ? ? 𝑦 𝑥 −5 5 ? ? ? 𝑦 𝑥
−2 2
2 2

13. Tangents and Normals of Hyperbolas
Find the equation of the tangent to the hyperbola with equation 𝑥 2 9 − 𝑦 2 4 =1 at the point 6,2 3 ?
2 9 𝑥− 2 4 𝑦 𝑑𝑦 𝑑𝑥 =1 At 6, , − 𝑑𝑦 𝑑𝑥 =1
→ 𝑚=
𝑦−2 3 = 𝑥−6
Bromusings: Pretty much the same as before! We could derive the parametric equations and differentiate these, but probably easier to differentiate Cartesian equation implicitly.
Prove that the equation of a tangent to the hyperbola 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1 at the point 𝑎 cosh 𝑡 , 𝑏 sinh 𝑡 is 𝑎𝑦 sinh 𝑡 +𝑎𝑏=𝑏𝑥 cosh 𝑡 ?
In Chapter 3 we’ll see 𝑑 𝑑𝑥 sinh 𝑥 = cosh 𝑥 and 𝑑 𝑑𝑥 cosh 𝑥 = sinh 𝑥
𝑑𝑦 𝑑𝑥 = 𝑑𝑦/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 𝑏 cosh 𝑡 𝑎 sinh 𝑡 𝑦−𝑏 sinh 𝑡 = 𝑏 cosh 𝑡 𝑎 sinh 𝑡 𝑥−𝑎 cosh 𝑡 …
Bromusings: Parametric equations given this time, so we could use parametric differentiation.

14. Test Your Understanding? ? ?

15. Exercise 2D

16. Foci and Directrices parabola
Recall from FP1 that the definition of a parabola is the locus of points equidistant from a point and a straight line.
We called the point the focus and the line the directrix.
directrix
focus

17. Click to Brosketch >
Foci and Directrices
But we could also have some fixed ratio between the distance to the focus and the distance to the directrix:
! Eccentricity 𝑒= 𝑃𝑆 𝑃𝐷
When 𝑒=1, we clearly get a parabola.
Click to Brosketch >
TASK: By plotting various points, determine what happens when 𝑒= 1 2
𝑒=1
! When 0<𝑒<1, the point 𝑃 describes an ellipse.
𝐷 2
𝐷 1
𝑒= 1 2
Because of the symmetry of an ellipse, we could have equivalently put the focus and directrix on the other side.
𝑆 1
𝑆 2 𝑃

18. Click to Brosketch >
Foci and Directrices
TASK: By plotting various points, determine what happens when 𝑒=2
Eccentricity 𝑒= 𝑃𝑆 𝑃𝐷
Because of the symmetry of a hyperbola, we could have equivalently put the focus and directrix on the other side.
! When 𝑒>1, the point 𝑃 describes an hyperbola.
Click to Brosketch >
𝑒=1
𝐷 3
𝐷 2
𝐷 1
𝑆 3
𝑒=2
𝑆 1
𝑆 2 𝑃

19. Determining Eccentricity of an Ellipse
Show that for 0<𝑒<1, the ellipse with focus S=(𝑎𝑒,0) and directrix 𝑥= 𝑎 𝑒 has equation 𝑥 2 𝑎 𝑦 2 𝑏 2 =1 and find 𝑏.
As with FP1 where we ‘proved’ the equation of a parabola, we can similarly reason about distances here. 𝑦 ? 𝑃 𝐷
𝑃 𝑆 2 = 𝑥−𝑎𝑒 2 + 𝑦 2 𝑃 𝑀 2 = 𝑎 𝑒 −𝑥 2 = 𝑎−𝑒𝑥 2 𝑒 2
Since 𝑃𝑆 𝑃𝐷 =𝑒, 𝑃 𝑆 2 = 𝑒 2 𝑃 𝑀 2 gives:
… 𝑥 2 𝑎 𝑦 2 𝑎 2 1− 𝑒 2 =1 𝑥 𝑆
𝑥= 𝑎 𝑒 𝑥 𝑦 𝑏
The major axis is the 𝑥 axis provided that 𝑎>𝑏 (the minor axis it the 𝑦 axis)
! Ellipse with equation 𝑥 2 𝑎 𝑦 2 𝑏 2 =1 where 𝑎>𝑏:
Eccentricity 0<𝑒<1 found using 𝑏 2 = 𝑎 2 1− 𝑒 2
Foci: ±𝑎𝑒, Directrices: 𝑥=± 𝑎 𝑒 𝑎

20. Finding directrices/foci
Find foci of the ellipse with equation 𝑥 𝑦 2 4 =1 and give the equation of the directrices. Hence sketch the ellipse.
Find foci of the ellipse with equation 𝑥 𝑦 =1 and give the equation of the directrices. Hence sketch the ellipse.
𝑎=3, 𝑏=2 𝑏 2 = 𝑎 2 1− 𝑒 → 9=4 1− 𝑒 2 ∴𝑒=
Foci: ±𝑎𝑒,0 = (± 5 , 0)
Directrices: 𝑥=± 𝑎 𝑒 =± 9 5 ? ?
𝑎=4, 𝑏=5
Because 𝑏>𝑎: 𝑎 2 = 𝑏 2 1− 𝑒 2 → 16=25 1− 𝑒 2 ∴𝑒= 3 5
Foci: 0,±𝑏𝑒 = 0,±3
Directrices: 𝑦=± 25 3 𝑦 2
𝑥= 9 5 𝑆′ 𝑆 𝑥 −3 3 𝑦
𝑦= 25 3
𝑥=− 9 5 5 −2
It makes sense that if we ‘rotate’ the ellipse, we also rotate the foci and directrices. 𝑆 𝑥
Ellipse with equation 𝑥 2 𝑎 𝑦 2 𝑏 2 =1 where 𝑎>𝑏:
Eccentricity 0<𝑒<1 found using 𝑏 2 = 𝑎 2 1− 𝑒 2
Foci: ±𝑎𝑒, Directrices: 𝑥=± 𝑎 𝑒 −4 4 𝑆′
𝑦=− 25 3 5

21. Test Your Understanding?

22. More on Ellipses
If 𝑃 is a point on an ellipse 𝑥 2 𝑎 𝑦 2 𝑏 2 =1, prove that 𝑃𝑆+𝑃 𝑆 ′ =2𝑎
i.e. That the width of the ellipse is the same as the width of the line formed by connecting the two foci to any point on the ellipse.
(Hint: Find a length for 𝑃𝐷 and 𝑃𝐷′ first) 𝑦 𝑥 𝑆′ 𝑎 𝑆
𝑥= 𝑎 𝑒
𝑥=− 𝑎 𝑒 −𝑎
𝑃(𝑥,𝑦) 𝐷 𝐷′ ?
𝑃 𝐷 ′ =𝑥+ 𝑎 𝑒 𝑃𝐷= 𝑎 𝑒 −𝑥 ∴𝑃𝑆+𝑃 𝑆 ′ =𝑒𝑃𝐷+𝑒𝑃 𝐷 ′ =𝑒 𝑎 𝑒 −𝑥 +𝑒 𝑎 𝑒 +𝑥 =𝑎−𝑒𝑥+𝑎+𝑒𝑥 =2𝑎

23. Back to Hyperbolas… 𝑥 𝑦
Show that for 𝑒>1, the hyperbola with foci at ±𝑎𝑒,0 and directrices at 𝑥=± 𝑎 𝑒 has equation 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1
𝑥=− 𝑎 𝑒
𝑥= 𝑎 𝑒
𝑃(𝑥,𝑦) 𝐷 ?
𝑃𝑆 𝑃𝐷 =𝑒 𝑃𝑆= 𝑥−𝑎𝑒 2 + 𝑦 2 𝑃𝐷=𝑥−𝑎𝑒 𝑃 𝑆 2 = 𝑒 2 𝑃 𝐷 2 … 1= 𝑥 2 𝑎 2 − 𝑦 2 𝑎 2 𝑒 2 −1
If 𝑏 2 = 𝑎 2 𝑒 2 −1 we obtain the standard equation of a hyperbola. 𝑆
−𝑎𝑒 −𝑎 𝑎 𝑎𝑒
Bro Notes: Note that the foci and directrices are EXACTLY THE SAME as ellipses.
However, because 𝑒>1, we’ll find that 𝑎𝑒>𝑎, i.e. the positive focus appears right of the curve, whereas with an ellipse it appears to the left. And vice versa with the directrices.
Note this is identical to the proof for ellipses: previously 𝑏 2 = 𝑎 2 1− 𝑒 2 which meant we had + 𝑦 2 𝑏 2 instead of − 𝑦 2 𝑏 2

24. Sketching Hyperbolas
Bro Note: For hyperbolas, you don’t care which of 𝑎 and 𝑏 are bigger. For ellipses, swapping the 𝑎 and 𝑏 has the effect of rotating the ellipse 90° and hence the foci/directrices too. We don’t get this same rotation for hyperbolas.
Sketch the hyperbola with equation 𝑥 2 9 − 𝑦 2 4 =1, indicating the foci and equations of the asymptotes.
Foci are ±𝑎𝑒,0 and directrices 𝑥=± 𝑎 𝑒 , but we need 𝑎 and 𝑒 first. Use 𝑏 2 = 𝑎 2 𝑒 2 −1
𝑎=3, 𝑏=2 4=9 𝑒 2 −1 → 𝑒= Foci: (± 𝟏𝟑 , 𝟎)
Asymptotes: 𝒚=+ 𝟐 𝟑 𝒙
(Formula booklet) ? 3 −3
𝑦=− 2 3 𝑥
𝑦= 2 3 𝑥 𝐷 𝑆 𝐷′ 13
− 13 𝑆′

25. Exercise 2E 1
Find the eccentricity of the following ellipses:
𝑥 𝑦 2 5 = → 𝑒= 𝑥 𝑦 2 9 = → 𝑒= 𝑥 𝑦 2 8 = → 𝑒= Find the foci and directrices of the following ellipses:
𝑥 𝑦 2 3 = → 𝑆 ±1,0 𝑥=±4 𝑥 𝑦 2 7 = →𝑆 ±3,0 𝑦=± 𝑥 𝑦 2 9 = →𝑆 0,±2 𝑦=± 9 2
An ellipse 𝐸 has focus 3,0 and the equation of the directrix is 𝑥=12. Find the value of the eccentricity and the equation of the ellipse.
𝑒= 1 2 , 𝑥 𝑦 =1 ? ? ? 2 ? ? ? 3 ?

26. Exercise 2E
An ellipse 𝐸 has focus 𝑞,0 and the equation of the directrix is 𝑥=8. Find the value of the eccentricity and the equation of the ellipse.
𝒆= 𝟏 𝟐 , 𝒙 𝟐 𝟏𝟔 + 𝒚 𝟐 𝟏𝟐 =𝟏
Find the eccentricity of the following hyperbolae:
𝑥 2 5 − 𝑦 2 3 =1, 𝑒= 𝟐 𝟏𝟎 𝟓 𝑥 2 9 − 𝑦 2 7 =1, 𝑒= 𝟒 𝟑 𝑥 2 9 − 𝑦 =1, 𝑒= 𝟓 𝟑 Find the foci of the following hyperbolae and sketch them, showing clearly the equations of the asymptotes.
𝑥 2 4 − 𝑦 2 8 =1, 𝒚=± 𝟐 𝒙, 𝒙−𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒔: ±𝟐,𝟎 𝑥 2 5 − 𝑦 2 3 =1, 𝒚=± 𝟑 𝟒 𝒙, 𝒙−𝒊𝒏𝒕𝒆𝒓𝒄𝒆𝒑𝒕𝒔: ±𝟒,𝟎 4 ? 5 ? ? ? 6 ? ?

27. Quick side note… 𝑦
𝑦= 1 𝑥 !
𝑆 2 , 2
Do you remember back in FP1 we had ‘rectangular hyperbolae’, which just seemed to be a fancy name for ‘reciprocal graphs’ (like 𝑦= 1 𝑥 ) that you had seen at GCSE? 𝑥 𝐷
𝑆′ − 2 ,− 2 𝐷′
A rectangular hyperbola is a hyperbola whose asymptotes are vertical and horizontal.
It’s possible to find the eccentricity, foci and directrices just as with a standard hyperbola.
The equation is NOT of the form 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1 because a standard hyperbola’s foci are on the 𝑥 axis.
Source: WolframAlpha

28. ? Get Cartesian equation
Loci questions
Recall that the locus of points are the set of points that satisfy some constraint.
The tangent to the ellipse with equation 𝑥 2 𝑎 𝑦 2 𝑏 2 =1 at the point 𝑃 𝑎 cos 𝑡 ,𝑏 sin 𝑡 crosses the 𝑥-axis at 𝐴 and the 𝑦-axis at 𝐵.
Find an equation for the locus of the mid-point of 𝐴𝐵 as 𝑃 moves around the ellipse.
Bro Tip: Draw a sketch! 𝑦
? Sketch 𝐵
? Get Cartesian equation 𝑀
From C4 you know how to find Cartesian equations from parametric ones.
cos 𝑡 = 𝑎 2𝑋 , sin 𝑡 = 𝑏 2𝑌 𝑎 2𝑋 𝑏 2𝑌 2 =1 𝑥 𝐴
What would you guess the shape of the locus 𝑀 makes as it moves?
? Tangent
𝑑𝑦 𝑑𝑥 = 𝑏 cos 𝑡 −𝑎 sin 𝑡
Equation of tangent:
𝑦−𝑏 sin 𝑡 = 𝑏 cos 𝑡 −𝑎 sin 𝑡 𝑥−𝑎 cos 𝑡 𝑎𝑦 sin 𝑡 +𝑏𝑥 cos 𝑡 =𝑎𝑏
Using 𝑦=0: 𝐴(𝑎 sec 𝑡 , 0) Similarly: 𝐵(0, 𝑏 𝑐𝑜𝑠𝑒𝑐 𝑡)
Midpoint: (𝑋,𝑌) where 𝑋= 𝑎 sec 𝑡 2 , 𝑌= 𝑏 cosec 𝑡 2
? Midpoint
Using capital 𝑋,𝑌 for this midpoint distinguishes from points on the original ellipse 𝑥,𝑦

29. Loci questions
The normal at 𝑃 𝑎 𝑝 2 , 2𝑎𝑝 and the normal at 𝑄 𝑎 𝑞 2 , 2𝑎𝑞 to the parabola with equation 𝑦 2 =4𝑎𝑥 meet at 𝑅.
(a) Find the coordinates of 𝑅.
The chord 𝑃𝑄 passes through the focus 𝑎,0 of the parabola.
(b) Show that 𝑝𝑞=−1
(c) Show that the locus of 𝑅 is a parabola with equation 𝑦 2 =𝑎 𝑥−3𝑎 ?
Normal at 𝑃: 𝑦+𝑝𝑥=2𝑎𝑝+𝑎 𝑝 3
Normal at 𝑄: 𝑦+𝑞𝑥=2𝑎𝑞+𝑎 𝑞 3
Solving simultaneously by eliminating 𝑥:
𝑥=2𝑎+𝑎 𝑝 2 +𝑝𝑞+ 𝑞 2 𝑦=−𝑎𝑝𝑞 𝑝+𝑞
Equation of chord:
𝑦 𝑝+𝑞 =2𝑥+2𝑎𝑝𝑞
Substituting in 𝑎,0 :
𝑝𝑞=−1
Bro Tip: Loci questions could concern any of the conics in either FP1 or FP3. a ? ? b ? ?
Tangent:
𝑡𝑦=𝑥+𝑎 𝑡 2
𝑥+ 𝑡 2 𝑦=2𝑐𝑡
You’re not given these. You could derive them, but it’s an advantage to memorise them.
Normal:
𝑦+𝑡𝑥=2𝑎𝑡+𝑎 𝑡 3
𝑡 3 𝑥−𝑡𝑦=𝑐 𝑡 4 −1

30. ? Get a single Cartesian equation.
Loci questions
The normal at 𝑃 𝑎 𝑝 2 , 2𝑎𝑝 and the normal at 𝑄 𝑎 𝑞 2 , 2𝑎𝑞 to the parabola with equation 𝑦 2 =4𝑎𝑥 meet at 𝑅.
(a) Find the coordinates of 𝑅.
The chord 𝑃𝑄 passes through the focus 𝑎,0 of the parabola.
(b) Show that 𝑝𝑞=−1
(c) Show that the locus of 𝑅 is a parabola with equation 𝑦 2 =𝑎 𝑥−3𝑎
Earlier we found 𝑅:
𝑥=2𝑎+𝑎 𝑝 2 +𝑝𝑞+ 𝑞 2 𝑦=−𝑎𝑝𝑞 𝑝+𝑞 c
Using 𝑝𝑞=−1, 𝑅 becomes:
𝑎+𝑎 𝑝 2 + 𝑞 2 , 𝑎 𝑝+𝑞
𝑋=𝑎+𝑎 𝑝 2 + 𝑞 2 𝑌=𝑎 𝑝+𝑞
𝑋=𝑎+𝑎 𝑎+𝑏 2 −2𝑎 =3𝑎+𝑎 𝑝+𝑞 2
=3𝑎+𝑎 𝑌 𝑎 2
Rearranging: 𝑌 2 =𝑎 𝑋−3𝑎 ?
Important Brogebra Tip:
𝑎 2 , 𝑏 2 , 𝑎𝑏 are related by 𝑎+𝑏 2 = 𝑎 2 + 𝑏 2 +2𝑎𝑏
We can use this identity to get 𝑎𝑏 into the equation whenever we see 𝑎 2 + 𝑏 2 and if we know what 𝑎𝑏 is.
? Get a single Cartesian equation.

31. Test Your Understanding? ? ?

32. Exercise 2F
The tangent at 𝑃 𝑎 𝑝 2 ,2𝑎𝑝 and the tangent at 𝑄 𝑎 𝑞 2 ,2𝑎𝑞 to the parabola with equation 𝑦 2 =4𝑎𝑥 meets at 𝑅.
(a) Find the coordinates of 𝑅. 𝒂𝒑𝒒, 𝒂 𝒑+𝒒
The chord 𝑃𝑄 passes through the focus 𝑎,0 of the parabola.
(b) Show that the locus of 𝑅 is the line 𝑥=−𝑎.
(c) Given instead that the chord 𝑃𝑄 has gradient 2, find the locus of 𝑅. 𝒚=𝒂
The tangent at 𝑃 𝑎 sec 𝑡 ,𝑏 tan 𝑡 to the hyperbola with equation 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1 cuts the 𝑥-axis at 𝐴 and the 𝑦-axis at 𝐵. Find the locus of the mid-point of 𝐴𝐵.
𝒂 𝟐 𝟒 𝒙 𝟐 =𝟏+ 𝒃 𝟐 𝟒 𝒚 𝟐
The normal at 𝑃 𝑎 sec 𝑡 ,𝑏 tan 𝑡 to the hyperbola with equation 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1 cuts the 𝑥-axis at 𝐴 and the 𝑦-axis at 𝐵. Find the locus of the mid-point of 𝐴𝐵.
𝟒 𝒂 𝟐 𝒙 𝟐 = 𝒂 𝟐 + 𝒃 𝟐 𝟐 +𝟒 𝒃 𝟐 𝒚 𝟐
The normal at 𝑃 𝑎 sin 𝑡 ,𝑏 cos 𝑡 to the hyperbola with ellipse 𝑥 2 𝑎 2 − 𝑦 2 𝑏 2 =1 cuts the 𝑥-axis at 𝐴 and the 𝑦-axis at 𝐵. Find the locus of the mid-point of 𝐴𝐵.
𝒙+ 𝒑 𝟐 𝒚=𝟐𝒄𝒑 1 ? ? 2 ? 3 ? 4 ?

33. Exercise 2F
The tangent from the point 𝑃 𝑐𝑝, 𝑐 𝑝 and the tangent from the point 𝑄 𝑐𝑞, 𝑐 𝑞 to the rectangular hyperbola 𝑥𝑦= 𝑐 2 , interest at the point 𝑅.
Show that 𝑅 is 2𝑐𝑝𝑞 𝑝+𝑞 , 2𝑐 𝑝+𝑞
Show that the chord 𝑃𝑄 has equation 𝑦𝑝𝑞+𝑥=𝑐 𝑝+𝑞
Find the locus of 𝑅 in the following cases (i) when the chord 𝑃𝑄 has gradient 2. 𝒚=−𝟐𝒙 (ii) when the chord 𝑃𝑄 passes through the point 1,0 𝒚=𝟐 𝒄 𝟐 (iii) when the chord 𝑃𝑄 passes through the point 0, 𝒙=𝟐 𝒄 𝟐
The chord 𝑃𝑄 to the rectangular hyperbola 𝑥𝑦= 𝑐 2 passes through the point 0,1 . Find the locus of the mid-point of 𝑃𝑄 as 𝑃 and 𝑄 vary.
𝒚= 𝟏 𝟐 5 ? ? ? 6 ?