Free Pearson Edexcel International A-level Mathematics · FP3

Ellipses and hyperbolas

What you’ll learn

  • Use eccentricity, foci, directrices and parameterisations.
  • Find conic tangents and normals.

Understand the idea

For an ellipse with major axis on the x-axis, a>b>0a>b>0 and the sum of focal distances is 2a2a. Its eccentricity is between 00 and 11. A horizontal hyperbola has eccentricity greater than 11, foci on the x-axis and asymptotes y=±bx/ay=\pm bx/a.

The parameterisation (acos⁡t,bsin⁡t)(a\cos t,b\sin t) traces an ellipse. For a hyperbola, (asec⁡t,btan⁡t)(a\sec t,b\tan t) covers both branches over suitable intervals; (acosh⁡t,bsinh⁡t)(a\cosh t,b\sinh t) covers the right branch. Implicit or parametric differentiation gives tangents and normals. A line is tangent when its substituted intersection quadratic has a repeated root.

Key formulae

x2a2+y2b2=1,e2=1−b2a2\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad e^2=1-\frac{b^2}{a^2}

x2a2−y2b2=1,e2=1+b2a2\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\quad e^2=1+\frac{b^2}{a^2}

Foci (±ae,0),directrices x=±a/e\text{Foci }(\pm ae,0),\qquad\text{directrices }x=\pm a/e

xx1a2±yy1b2=1(tangent at (x1,y1))\frac{xx_1}{a^2}\pm\frac{yy_1}{b^2}=1\quad\text{(tangent at }(x_1,y_1)\text{)}

Ellipse semiaxes determine intercepts and focal distance

Conditions. Ellipse x²/9+y²/4=1, parametrised by x=3 cos t,y=2 sin t. Equal coordinate-unit scales preserve its geometry.

x / coordinate units; y / coordinate units.

Horizontal intercepts are±3 and vertical intercepts±2. Focal distance from centre is√(9−4)=√5≈2.236, so the foci lie inside the horizontal vertices, not at the intercepts.

Open the interactive lesson to view plotted curves and coordinate tables.

Worked example

Find the eccentricity and foci of x2/25+y2/9=1x^2/25+y^2/9=1.

  1. The semi-axes are a=5a=5, b=3b=3.
  2. Thus e2=1−9/25=16/25e^2=1-9/25=16/25 and e=4/5e=4/5.
  3. The focal distance from the origin is ae=4ae=4.

Answer: e=4/5e=4/5; foci (4,0)(4,0) and (−4,0)(-4,0).

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