SHS3 Additional Mathematics · Semester 1, Week 16

Spatial Reasoning

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Curriculum details

Strand: Geometric Reasoning and Measurement (Strand 2)
Sub-strand: Spatial Reasoning (2.1)
Content standard: 3.2.1.CS.1 - Demonstrate an understanding of Parabola and its properties. 3.2.1.LO.1 Construct a parabola of a given quadratic equation (𝑦 = 𝑎𝑥 ) + 𝑏 + 𝑐) and explain its key features 3.2.1.LO.2 Sketch a parabola and use it to deduce the relation 𝑦 ) = 4𝑎 3.2.1.LO.3 Sketch a parabola given the directrix Communication: Provide learners the opportunity to engage and and focus. 3.2.1.LO.4 Deduce the equation of the tangent and normal to a parabola
Indicator: 3.2.1.LI.10 - Determine the equations of tangent and normal to a parabola.
Suggested placement: Semester 1, Week 16 (Week 16 of the year)
Our suggestion, laid out in curriculum order across three terms of twelve weeks. NaCCA does not fix the week, so follow your school's scheme of learning.
Source document note: The official curriculum document has a fault in this entry, so the curriculum text above is transcribed exactly as printed:

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Curriculum reference: NaCCA curriculum document, p. 493

Exemplars (from the NaCCA curriculum)

Experiential Learning, Talk for Learning, Group work, and Building on what others say.
Learning Experience: Learners in mixed-ability groups investigate the equation of the tangent and the equation of the normal to the parabola
Activity 1: Learners in their mixed-ability groups apply their knowledge in finding derivatives and the
equation of a line to investigate how to derive the equation of a tangent to the parabola 𝑦 ) = 4𝑎 at a point (𝑥 ! , 𝑦 ! ).
- <A The slope of a tangent is given by <0 ∴ differentiating 𝑦 ) = 4𝑎 on both sides with respect to 𝑥 will be 𝑑 2𝑦 = 4𝑎 𝑑 𝑑 2𝑎 = 𝑑 𝑦
- Therefore, the slope (𝑚) of the tangent to the parabola at the point (𝑥 ! , 𝑦 ! ) on it is )& A $ Now, the equation of the tangent to the parabola at the point (𝑥 ! , 𝑦 ! ) on it is 𝑦 − 𝑦 ! = 𝑚(𝑥 − 𝑥 ! ) 2𝑎 ⇒ 𝑦 − 𝑦 ! = (𝑥 − 𝑥 ! ) 𝑦 ! ⟹ 𝑦 ! − 𝑦 !) = 2𝑎 − 2𝑎𝑥 ! But 𝑦 ! = 4𝑎𝑥 ! ) 𝑦 ! − 4𝑎𝑥 ! = 2𝑎 − 2𝑎𝑥 ! ∴ 𝑦 ! = 2𝑎(𝑥 + 𝑥 ! ) is the equation of the tangent at a point (𝑥 ! , 𝑦 ! ) on the parabola.
Activity 2: Equation of normal to a parabola. Learners in mixed-ability groups apply their knowledge in finding derivatives and the equation of a line to explore and investigate how to derive the equation of normal to the parabola 𝑦 ) = 4𝑎 at a point (𝑥 ! , 𝑦 ! ). A
- Learners establish that the slope (𝑚) of the normal is − )& since normal and tangent are
perpendicular to each other. ∴ The equation of the normal 𝑦 − 𝑦 ! = 𝑚(𝑥 − 𝑥 ! ) 𝑦 ! ⇒ 𝑦 − 𝑦 ! = − (𝑥 − 𝑥 ! ) 2𝑎 ⟹ 𝑦 ! 𝑥 − 2𝑎𝑦 ! − 𝑥 ! 𝑦 ! + 2𝑎 = 0 is the equation of the normal at a point (𝑥 ! , 𝑦 ! ) on the parabola.
Example: A Parabola 𝑦 ) = 8𝑥 passes through the point 𝑃(1, 2). Find:
- equation of the tangent
- equation of the normal to the parabola at the point P
Solution i. If 𝑦 ) = 8𝑥, then comparing it to 𝑦 ) = 4𝑎 𝑎 = 2 Equation of the tangent to the parabola is given as 𝑦 ! = 2𝑎(𝑥 + 𝑥 ! ) ⇒ 𝑦(2) = 2 ∙ 2(𝑥 + 1) 𝑦 − 2𝑥 − 2 = 0 ii. 𝑦 ! 𝑥 − 2𝑎𝑦 ! − 𝑥 ! 𝑦 ! + 2𝑎 = 0 ⟹ 2𝑥 − 2(2)2 − (1)(1) + 2(2)𝑦 = 0 2𝑥 + 4𝑦 − 9 = 0
Activity 3: Learners in groups create and pose problems for other groups to solve and share their solutions.
Teaching and Learning Resources:
- SHS Curriculum, Graph boards, mathematical set, ICT tools
Assessment (3.2.1.AS.10). 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.

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