Spatial Reasoning

Experiential Learning, Talk for Learning, Group work, and Building on what others say.
Learning Experience: Learners in mixed-ability groups deduce the directrix and focus of a parabola as well as write the equation of a parabola given its focus and directrix.
Activity 1 Deducing the directrix and focus of a given parabola with vertex at the origin. Learners in their mixed-ability group brainstorm to deduce the directrix and focus of a given parabola with vertex at the origin.
Example: Find the focus and directrix of 𝑦 ) = −8𝑥
Solution: Compare the equation given to the standard equation of a parabola, i.e. 𝑦 ) = 4𝑎 ∴ 4𝑎 = −8 𝑎 = −2 Therefore, the focus of the parabola is (0, −2), and the directrix is 2.
Activity 2: Learners in their mixed-ability groups create and solve questions on deducing the directrix and focus of a given parabola with vertex at the origin.
Activity 3 Deducing the directrix and focus of a given parabola of a parabola not at the origin. Learners in their mixed-ability groups to brainstorm to deduce the directrix and focus of a given parabola with the vertex not at the origin.
Example: Find the directrix and focus of (𝑦 − 1) ) = −16(𝑥 + 3) Solution
- Comparing the equation to (𝑦 − 𝑘) ) = 4𝑝(𝑥 − ℎ), ℎ = −3 𝑎 𝑘 = 1)
- Also 4𝑝 = −16, and 𝑝 = −4
- Since 𝑝 < 0, the parabola opens left, ∴ 𝑡ℎ𝑒 𝑐 𝑜 𝑡ℎ𝑒 𝑓 𝑖 (ℎ + 𝑝, 𝑘) ∴ 𝑓 = (−3 − 4,1) (−7,1)
Activity 4: Learners in their mixed-ability groups create and solve questions on deducing the directrix and focus of a given parabola with vertex not at the origin.
Activity 5: Deriving the equation of a parabola given its focus and directrix. Learners in their mixed-ability groups brainstorm on how to derive the equation of a parabola, given its focus and directrix.
Example: Determine the equation of the parabola with focus (3, −1) and directrix 𝑥 = 6.
Solution
- Learners led by a leader brainstorm to sketch the curve and deduce the general equation of this parabola as (𝑦 − 𝑘) ) = −4𝑎(𝑥 − ℎ)

- Through interaction and building on what others say, learners establish that the vertex (ℎ, 𝑘) is the midpoint of (3, −1) and (6, −1) 3+6 ∴ (ℎ, 𝑘) = £ , −1¤ 2 9 = £ , −1¤ 2
- Through interaction and building on what others say, learners establish that 𝑎 is half the distance /*" " between the focus and the directrix, that is ) = ) , since it opens to the left 𝑎 < 0
Therefore, the equation of the parabola is 3 9 (𝑦 + 1) ) = 4 ∙ − £ ¤ £𝑥 − ¤ 2 2 9 (𝑦 + 1) ) = −6 £𝑥 − ¤ 2
Example: Determine the equation of the parabola with focus (−2,13) and directrix 𝑦 = 5.
Solution: The equation of the parabola is (𝑥 − ℎ) ) = 4𝑎(𝑦 − 𝑘) The vertex is the midpoint of the focus and directrix; 13 + 5 = £−2, ¤ 2 = (−2,9)
𝑎 is half the distance between from (−2,13) to (2,5) 13 − 5 =4 2 ∴ 𝑡ℎ𝑒 𝑒 𝑖 (𝑥 + 2) ) = 4 ∙ 4(𝑦 − 9) (𝑥 + 2) ) = 16(𝑦 − 9)
Activity 5: Learners in their mixed-ability groups brainstorm to make generalisations on how to derive the equation of a parabola given its focus and directrix. NB: In general, the equation of the parabola with focus (𝑎, 𝑏) and directrix 𝑦 = 𝑐 is (𝑥 − 𝑎) ) + 𝑏 ) − 𝑐 ) = 2(𝑏 − 𝑐)𝑦
Teaching and Learning Resources:
- SHS Curriculum, Graph boards, mathematical set, ICT tools
Assessment (3.2.1.AS.6). The document marks these depth-of-knowledge levels for this indicator: Level 2 Skills of conceptual understanding; Level 3 Strategic reasoning; Level 4 Extended critical thinking and reasoning.