# 4.6 Hyperbola (*e* > 1)

A *hyperbola* is the set of points *P* in the plane whose distances from a fixed point *F* are *e* times their distances from a fixed line *d*, where *e* > 1. We obtain a hyperbola *in standard form* if

the focus

*F*lies on the*x*-axis, and has coordinates (*ae*, 0), where*a*> 0;the directrix

*d*is the line with equation*x*=*a*/*e*.

Let *P* (*x*, *y*) be an arbitrary point on the hyperbola, and let *M* be the foot of the perpendicular from *P* to the directrix. Since *FP* = *e* × *PM*, by the definition of the hyperbola, it follows that *FP*^{2} = *e*^{2} × *PM*^{2}; that is,

Multiplying out the brackets, we obtain

*x*^{2} − 2*aex* + *a*_{2}*e*_{2} + *y*_{2} = *e*_{2}*x*_{2} − 2*aex* + *a*_{2},

which simplifies to the equation

*x*_{2}(*e*_{2} − 1) − *y*_{2} = *a*_{2}(*e*_{2} − 1),

that is,

Substituting *b* for , so that *b*^{2} = *a*^{2}(*e*^{2} − 1), we obtain the standard form of the equation of the hyperbola

This equation is symmetric in *x* and in *y*, so that the hyperbola also has a second focus *F*′ at (−*ae*, 0), and a second directrix *d*′ with equation *x* = −*a*/*e*.

The hyperbola intersects the *x*-axis at the points (±*a*, 0). We call the line segment joining the points (±*a*, 0) the *major axis* or *transverse axis* of the hyperbola, and the line segment joining the points (0, ±*b*) the *minor axis* or *conjugate axis* of the hyperbola (this is not a chord of the hyperbola). The origin is the centre of this hyperbola.

Each point with coordinates (*a* sec *t*, *b* tan *t*) lies on the hyperbola, since

*Note: * In general, sec^{2}*t* = 1 + tan^{2}*t*.

Then, just as for the parabola, we can check that

gives a parametric representation of the hyperbola.

*Note:* An alternative parametrisation, using hyperbolic functions, is *x* = *a*cosh*t*, y = *b*sinh*t* (*t* ∈ ).

Two other features of the shape of the hyperbola stand out.

First, the hyperbola consists of two separate curves or *branches*.

Secondly, the lines with equations *y* = ±*bx*/*a* divide the plane into two pairs of opposite sectors; the branches of the hyperbola lie in one pair. As *x* → ±∞, the branches of the hyperbola get closer and closer to these two lines. We call the lines *y* = ±*bx*/*a* the *asymptotes* of the hyperbola.

We summarise these facts as follows:

## Hyperbola in standard form

A hyperbola in standard form has equation

It can also be described by the parametric equations

It has foci (±*ae*, 0) and directrices *x* = ±*a*/*e*; its major axis is the line segment joining the points (±*a*, 0), and its minor axis is the line segment joining the points (0, ±*b*).

## Example 57

Let *P* be a point , *t* ∈ , on the hyperbola with equation *x*^{2} − 2*y*^{2} = 1.

**(a)**Determine the foci*F*and*F*′ of the hyperbola.**(b)**Determine the gradients of*FP*and*F*′*P*, when these lines are not parallel to the*y*-axis.**(c)**Find the point*P*on the hyperbola, in the first quadrant, for which*FP*is perpendicular to*F*′*P*.

### Answer

**(a)**This hyperbola is of the form with*a*= 1 and , so . If*e*denotes the eccentricity of the hyperbola, so that*b*^{2}=*a*^{2}(*e*^{2}− 1), we haveit follows that , so

In the general case, the foci are (±

*ae*, 0); it follows that here the foci are .**(b)**Let*F*and*F*′ be and , respectively. (It does not matter which way round these are chosen.)Then the gradient of

*FP*iswhere we know that , since

*FP*is not parallel to the*y*-axis.Similarly, the gradient of

*F*′*P*iswhere we know that , since

*F*′*P*is not parallel to the*y*-axis.**(c)**When*FP*is perpendicular to*F*′*P*, we haveWe may rewrite this in the form

so 2sec

^{2}*t*− 3 + tan^{2}*t*= 0.Since sec

^{2}*t*= 1 + tan^{2}*t*, it follows that 3 tan^{2}*t*= 1.Since we are looking for a point

*P*in the first quadrant, we choose .When , we have . Since we are looking for a point

*P*in the first quadrant, we choose .It follows that the required point

*P*has coordinates .