IP AMaths Notes (Upper Sec, Year 3-4): 07) Parabolas and Circles

Upper-sec AMaths emphasises the ability to flip between algebraic equations and geometric interpretations of parabolas and circles.

Standard forms

• Circle with centre \( (a, b) \) and radius \( r \): \( (x - a)^{2} + (y - b)^{2} = r^{2} \).
• Expanded circle: \( x^{2} + y^{2} + Dx + Ey + F = 0 \) with centre \( (-\tfrac{D}{2}, -\tfrac{E}{2}) \), radius \( r = \sqrt{\bigl(\tfrac{D}{2}\bigr)^{2} + \bigl(\tfrac{E}{2}\bigr)^{2} - F} \).
• Parabola opening right: \( (y - k)^{2} = 4p(x - h) \) with focus \( (h + p, k) \), directrix \( x = h - p \).
• Parabola opening up: \( (x - h)^{2} = 4p(y - k) \) with focus \( (h, k + p) \).

Worked example 1 — Circle from three points

Find the equation of the circle passing through \( (1, 2) \), \( (5, 4) \), and \( (3, -2) \).

1. Use general form \( x^{2} + y^{2} + Dx + Ey + F = 0 \).
2. Substitute each point to generate simultaneous equations:
   • \( 1 + 4 + D + 2E + F = 0 \) → \( D + 2E + F = -5 \).
   • \( 25 + 16 + 5D + 4E + F = 0 \) → \( 5D + 4E + F = -41 \).
   • \( 9 + 4 + 3D - 2E + F = 0 \) → \( 3D - 2E + F = -13 \).
3. Solve the linear system. Subtract first equation from second: \( 4D + 2E = -36 \) so \( 2D + E = -18 \).
4. Subtract first from third: \( 2D - 4E = -8 \) so \( D - 2E = -4 \).
5. Solve simultaneously: multiply \( D - 2E = -4 \) by 2 → \( 2D - 4E = -8 \).
6. Subtract from \( 2D + E = -18 \): \( 5E = -10 \) giving \( E = -2 \).
7. Substitute back: \( 2D - (-4) = -8 \) → \( 2D + 4 = -8 \) so \( D = -6 \).
8. Use first equation: \( -6 + 2(-2) + F = -5 \) → \( -10 + F = -5 \) giving \( F = 5 \).
9. Circle equation: \( x^{2} + y^{2} - 6x - 2y + 5 = 0 \).

Centre \( (3, 1) \); radius \( r = \sqrt{3^{2} + 1^{2} - 5} = \sqrt{5} \).

Worked example 2 — Tangent to a parabola

Find the equation of the tangent to \( y^{2} = 8x \) at the point where \( y = 4 \).

1. Substitute to get point: \( 4^{2} = 8x \) gives \( 16 = 8x \) so \( x = 2 \). Point is \( (2, 4) \).
2. Differentiate implicitly: \( 2y \dfrac{dy}{dx} = 8 \).
3. Thus \( \dfrac{dy}{dx} = \dfrac{8}{2y} = \dfrac{4}{y} \).
4. At \( y = 4 \), gradient \( m = 1 \).
5. Equation: \( y - 4 = 1(x - 2) \) ⇒ \( y = x + 2 \).

Try this

A parabola has equation \( (x - 1)^{2} = 12(y + 2) \). Determine its focus, directrix, and the equation of the tangent at \( (4, 4) \).