Two Dimensional Analytical Geometry
This chapter moves from locus to straight lines, pairs of straight lines, circles and the three standard conics. The exam favours finding an equation from given data and reading centre, radius, focus, directrix or eccentricity from a standard form.
Straight lines
Slope form through (x₁, y₁): y − y₁ = m(x − x₁); intercept form x/a + y/b = 1. The line ax + by + c = 0 has slope −a/b. The perpendicular distance from (x₁, y₁) to that line is |ax₁ + by₁ + c|/√(a² + b²). Parallel lines share a slope; perpendicular lines satisfy m₁m₂ = −1. The angle between two lines satisfies tan θ = |(m₁ − m₂)/(1 + m₁m₂)|.
Circles and conics
The circle (x − h)² + (y − k)² = r² expands to x² + y² + 2gx + 2fy + c = 0, with centre (−g, −f) and radius √(g² + f² − c). The parabola y² = 4ax has focus (a, 0) and directrix x = −a. For the ellipse x²/a² + y²/b² = 1 with a > b, e² = 1 − b²/a²; for the hyperbola x²/a² − y²/b² = 1, e² = 1 + b²/a².
Worked example
Find the circle with centre (2, −1) passing through (5, 3). The radius squared is (5 − 2)² + (3 + 1)² = 9 + 16 = 25, so r = 5. The equation is (x − 2)² + (y + 1)² = 25, that is x² + y² − 4x + 2y − 20 = 0.
Traps
- Omitting the modulus in the perpendicular distance formula.
- Using the ellipse eccentricity relation for the hyperbola.
- Forgetting to square the radius in the circle equation.