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Mathematics · Ch 5 — Two Dimensional Analytical Geometry-II

Hyperbola

5.3.4

Hyperbola

A hyperbola is the locus with e>1e>1: its distance from the focus is greater than its distance from the directrix, scaled by ee.

(i) Standard form. With the same construction as the ellipse (points A,A′A,A' dividing SZSZ internally/externally in ratio e:1e:1, AA′=2aAA'=2a, CC the midpoint as origin) but now e>1e>1, working through AS=eAZAS=eAZ, A′S=eA′ZA'S=eA'Z gives again CS=aeCS=ae and CZ=a/eCZ=a/e, so focus S(ae,0)S(ae,0), directrix x=a/ex=a/e. Applying SP2=e2PM2SP^2=e^2PM^2 and simplifying, with b2=a2(e2−1)b^2=a^2(e^2-1) (positive since e>1e>1), gives

x2a2−y2b2=1,\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,

the hyperbola in standard form, symmetric about both axes, with a second focus S′(−ae,0)S'(-ae,0) and matching directrix x=−a/ex=-a/e by symmetry, exactly as for the ellipse.

Vocabulary (Definition 5.5). AA′AA' (length 2a2a, A(a,0),A′(−a,0)A(a,0),A'(-a,0) the two vertices) is the transverse axis; BB′BB' (length 2b2b, where B(0,b)B(0,b) is not on the curve) is the conjugate axis. Taking ae=cae=c, so b2=c2−a2b^2=c^2-a^2: foci (±c,0)(\pm c,0), directrices x=±a/ex=\pm a/e, and (by the same substitution as for the ellipse) latus rectum =2b2/a=2b^2/a — proved directly in Ex. 5.2 Q6.

Key difference-of-focal-distances property (Ex. 5.2 Q7). For any point PP on the hyperbola, ∣PS−PS′∣=2a|PS-PS'|=2a (constant) — the hyperbola's analogue of Theorem 5.5, proved the same way from PS=ex−aPS=ex-a, PS′=ex+aPS'=ex+a on the right branch (and symmetrically on the left).

Asymptotes. As a point on the curve moves further from the centre, the hyperbola's two branches approach, but never touch, two straight lines called asymptotes — a feature the parabola and ellipse do not have. The circle described on the transverse axis as diameter, x2+y2=a2x^2+y^2=a^2, is again called the auxiliary circle (used for parametrising the hyperbola, §5.5.1). …

Figure 5.27–5.28Parts of a hyperbola

What this figure shows. Both branches with centre CC, both foci S,S′S,S', both vertices A,A′A,A' and the latus rectum LL′LL', for x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1. …

Figure 5.29–5.30Hyperbola with centre (h,k)

What this figure shows. Two orientations of a shifted hyperbola — transverse axis parallel to the xx-axis and parallel to the yy-axis — each labelled with its shifted centre, vertices and foci. …