Free Pearson Edexcel International A-level Mathematics · FP1

Parabolas and rectangular hyperbolas

What you’ll learn

  • Use Cartesian and parametric forms of basic conics.
  • Find tangents, normals and intersections.

Understand the idea

The parabola y2=4axy^2=4ax has focus (a,0)(a,0) and directrix x=−ax=-a. The parameterisation (at2,2at)(at^2,2at) places every real value of tt on the curve and often makes tangent calculations shorter.

A rectangular hyperbola xy=c2xy=c^2 has the coordinate axes as asymptotes and can be parameterised by (ct,c/t)(ct,c/t), where t≠0t\ne0. Differentiate implicitly or parametrically to obtain a gradient, then use the point-gradient equation; solve simultaneous equations to investigate a line's intersections with a conic.

Key formulae

y2=4ax,(x,y)=(at2,2at)y^2=4ax,\qquad(x,y)=(at^2,2at)

ty=x+at2ty=x+at^2

xy=c2,(x,y)=(ct,c/t)xy=c^2,\qquad(x,y)=(ct,c/t)

dydx=−yx(xy=c2)\frac{dy}{dx}=-\frac{y}{x}\quad(xy=c^2)

Worked example

Find the tangent to y2=8xy^2=8x at parameter t=3t=3.

  1. Here 4a=84a=8, so a=2a=2.
  2. The point is (at2,2at)=(18,12)(at^2,2at)=(18,12).
  3. The tangent formula gives 3y=x+2(32)3y=x+2(3^2).

Answer: x−3y+18=0x-3y+18=0.

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