Coordinate geometry and conic equations
Identify a conic's standard form, centre or vertex, scale and orientation.
What you will tackle
- Interpret coordinate equations and transformations of conics
- Identify circles, parabolas, ellipses and hyperbolas in standard forms
- Use graph features, point tests and distance relationships
Recognise each form
A circle uses (x − h)² + (y − k)² = r². A vertical parabola y − k = a(x − h)² has vertex (h,k) and opens up for a > 0 or down for a < 0. An ellipse (x − h)²/a² + (y − k)²/b² = 1 has semiaxes a and b. A hyperbola has a difference of squared terms; the positive term identifies the transverse-axis direction.
Use the equation to check the picture
The denominators in an ellipse or hyperbola are squared scale lengths. A translation changes h and k, not those scale lengths. Substituting a point tests membership. For a horizontal hyperbola x²/a² − y²/b² = 1, vertices are (±a,0) and asymptotes are y = ±(b/a)x. More elaborate conic theory is not needed to recognise these features.
Build your notes, then practise
Work on paper before opening an explanation. Guided tasks provide a starting point; independent tasks remove the scaffolding; application tasks ask you to justify a method or interpretation in context.
Guided practice
Guided 1
Identify the shape and main dimensions: (x − 1)² + (y + 2)² = 9; x²/25 + y²/9 = 1; y = −2(x − 3)² + 4.
Show a starting hint
Look for the signs and whether both variables are squared.
Compare with the explained answer
- The first is a circle with centre (1,−2), radius 3.
- The ellipse has centre (0,0), horizontal semiaxis 5 and vertical semiaxis 3.
- The parabola has vertex (3,4) and opens down.
Answer / response criteria: Circle, ellipse and downward-opening parabola, with features as stated.
Independent practice
Independent 1
For x²/16 − y²/9 = 1, find the vertices and asymptotes; test (4,0) and (0,3). Translate its centre to (2,−1).
Compare with the explained answer
- a = 4 and b = 3, so vertices are (±4,0) and asymptotes y = ±3x/4.
- (4,0) gives 1 and lies on it; (0,3) gives −1 and does not.
- The translated equation is (x − 2)²/16 − (y + 1)²/9 = 1.
Answer / response criteria: Vertices (±4,0); asymptotes y = ±3x/4; only (4,0) is on the original; translated equation as stated.
Application practice
Application 1
An elliptical garden is modelled by x²/36 + y²/16 = 1 in metres. Find its full width and height, test (0,4), and explain why substituting x = 6 gives y = 0.
Compare with the explained answer
- Semiaxes are 6 and 4, so full dimensions are 12 by 8.
- (0,4) gives 16/16 = 1.
- At x = 6 the first term is already 1; the remaining squared term must be zero.
Answer / response criteria: 12 m wide and 8 m high; (0,4) lies on it; y = 0 at x = 6.
What can you explain now?
An elliptical garden is modelled by x²/36 + y²/16 = 1 in metres. Find its full width and height, test (0,4), and explain why substituting x = 6 gives y = 0.
Compare with the explanation
12 m wide and 8 m high; (0,4) lies on it; y = 0 at x = 6.
Explain the reasoning before comparing. This review records your self-assessment rather than an official score or mastery decision.
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Source and review status
Coverage follows the named requirements and examples in the 2026–2027 ACT test guide and subject descriptions. That guide explicitly does not list every possible topic. The broader coded College and Career Readiness Standards are also mapped to conceptual teaching and practice; this does not calibrate a learner to a score band. The separate scientific-background context list and reading-complexity rubric have their own stated limits.
Original teaching and explained self-check practice are available. Specialist review and difficulty calibration remain pending. Essays require self-check or teacher review and are not automatically marked or awarded official scores.
Official source specification / guide