Conic Sections

1Focus–directrix definition and eccentricity

Every conic is the locus of points whose distance from a focus is e times the distance from the corresponding directrix. The value of e classifies the curve: parabola when e=1, ellipse when e<1, hyperbola when e>1 — so eccentricity is the single number that decides which standard form you open.

Figure. Parabola y^2=4ax (a=2): every point P on the curve is equidistant from the focus F(a,0) and the directrix x=-a, so e=1.

2Standard parabola y^2=4ax

The standard parabola y^2=4ax has vertex at the origin, focus (a,0), directrix x=-a and latus rectum of length 4a. Its eccentricity is e=1, and the parametric point is (at^2,2at). Read a as one quarter of the coefficient of x — that single reading gives focus and latus rectum together.

Figure. The parabola y^2=12x (a=3) opening right from the origin, with focus F(3,0) marked on the axis. Axes are part of the standard-form reading.

Reading y^2=4ax

1. Extract aCompare with y^2=4ax: the coefficient of x is 4a.
2. Focus and directrixFocus (a,0), directrix x=-a.
3. Latus rectumLength 4a — the same number as the coefficient of x.

3Standard ellipse and eccentricity

For \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 with a>b, the foci are (\pm ae,0) where b^2=a^2(1-e^2), so e<1. The sum of focal distances from any point on the ellipse is 2a, and the latus rectum is \dfrac{2b^2}{a}. Take the larger denominator as a squared — swapping the axes misplaces the foci.

Figure. Ellipse \dfrac{x^2}{25}+\dfrac{y^2}{9}=1 about centred axes, with foci at (\pm 4,0) marked. Sampled from the Cartesian equation in the plot frame.

Eccentricity from the axes

1. Identify a^2>b^2The larger denominator under x^2 (for a major axis along x) is a^2.
2. Compute ee=\sqrt{1-b^2/a^2}.
3. Place the fociFoci at (\pm ae,0) on the major axis.

4Standard hyperbola and asymptotes

For \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1, the foci are (\pm ae,0) with b^2=a^2(e^2-1), so e>1. The difference of focal distances is 2a, the asymptotes are y=\pm\dfrac{b}{a}x, and a rectangular hyperbola has eccentricity root two (equivalently a=b). Get eccentricity from e=\sqrt{1+b^2/a^2}.

Figure. Hyperbola x^2/a^2-y^2/b^2=1 with both branches and the asymptote y=(b/a)x; the partner y=-(b/a)x is the mirror. Eccentricity is e=\sqrt{1+b^2/a^2}>1.

Reading the hyperbola

1. Identify a,bTransverse axis length 2a sits under the positive term.
2. Eccentricitye=\sqrt{1+b^2/a^2}, always greater than 1.
3. AsymptotesDraw y=\pm (b/a)x — the branches approach these lines.

5Latus rectum across conics

The latus rectum of an ellipse or hyperbola equals \dfrac{2b^2}{a}; for the parabola y^2=4ax it is 4a. Use a given latus rectum as a chord length to read b or a directly — it is often the fastest route into the axes.

Figure. Latus rectum of y^2=4ax is the focal chord x=a, of length 4a (here a=3, length 12). For the ellipse it is 2b^2/a.

6Tangents with slope m

The tangent to y^2=4ax with slope m is y=mx+\dfrac{a}{m}; to the ellipse y=mx\pm\sqrt{a^2m^2+b^2}; to the hyperbola y=mx\pm\sqrt{a^2m^2-b^2}. Each form is the condition of tangency: substitute the line into the conic and set the discriminant to zero.

Figure. At (at^2,2at) on y^2=4ax the tangent is ty=x+at^2. Slope-m form y=mx+a/m is the same family for the parabola.

Condition of tangency

1. Write the lineTake y=mx+c (or the point–slope form the question gives).
2. Substitute into the conicReplace y and clear to a quadratic in x.
3. Set discriminant zeroOne intersection (tangent) forces D=0, which rearranges to the listed \pm\sqrt{\cdots} forms.

7Eccentricity shortcuts from a and b

Get eccentricity fast from the axes: ellipse e=\sqrt{1-\dfrac{b^2}{a^2}}, hyperbola e=\sqrt{1+\dfrac{b^2}{a^2}}, with a the semi-major or transverse semi-axis. Parabola is the boundary case e=1 with no second axis length to compare.

Same axes as the ellipse and parabola cards — only the formula for e changes sign between ellipse and hyperbola.