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Analytic geometry - Conic sections
 Hyperbola and line
 Hyperbola and line relationships

  Let examine relationships between a hyperbola and a line passing through the center of the hyperbola, i.e., the origin. A line y = mx intersects the hyperbola at two points if the slope

   |m| < b/a  but if  |m| > b/a  then, the line

  y = mx  does not intersect the hyperbola at all.

The diameters of a hyperbola are straight lines passing through its center.

The asymptotes divide these two pencils of diameters into, one which intersects the curve at two points, and  the other which do not intersect.

A diameter of a conic section is a line which passes through the midpoints of parallel chords.

Conjugate diameters of the hyperbola (or the ellipse) are two diameters such that each bisects all chords drawn parallel to the other.

As the equation of a hyperbola can be obtained from the equation of an ellipse by changing the sign of b2  
that is,     a2(1 - e2) = -a2(e2 - 1) = -b2,

this way, we can use other formulas relating to the ellipse to obtain corresponding formulas for the hyperbola. Therefore, to examine conditions which determine position of a line in relation to a hyperbola we solve the system of equations,

  y = mx + c                    then if,    a2m2 - b2 > c2   the line intersects the hyperbola at two points,
  b2x2 - a2y2 = a2b2                             a2m2 - b2 < c2   the line and the hyperbola do not intersect.
                                                     a2m2 - b2 = c2   the line is the tangent of the hyperbola,
Condition for a line to be the tangent to the hyperbola - tangency condition
A line is the tangent to the hyperbola if  a2m2 - b2 = c2,  this equality is called the tangency condition.

Regarding the asymptotes, to which c = 0, this condition gives |m| = b/a and that is why we can say that the hyperbola touches the asymptotes at infinity.

The tangency condition also states that the slopes of the tangents will satisfy the condition if
That is, the tangents to the hyperbola can only be parallel to a line belonging to the pencil of lines that do not intersects the hyperbola.
The equation of the tangent at the point on the hyperbola

The points of contact of a line and the hyperbola can be obtained from the corresponding formula for the ellipse by changing b2 with -b2 thus,

  the tangency point or the point of contact.
Thus, the intercept and slope of the tangent
   or     b2x1x - a2y1y = a2b2    the equation of the tangent at a point P1(x1, y1) on the hyperbola.
Polar and pole of the hyperbola

If from a point A(x0, y0), exterior to the hyperbola, drawn are tangents, then the secant line passing through the contact points, is the polar of the point A. The point A is called the pole of the polar.

The equation of the polar is derived the same way as for the ellipse,
b2x0x - a2y0y = a2b2      the equation of the polar of the point A(x0, y0).
 Construction of the tangent at the point on the hyperbola

The tangent at the point P1(x1, y1) on the hyperbola is the bisector of the angle F1P1F2 subtended by

focal radii, r1 and r2 at P1.

The proof shown for the ellipse applied to the hyperbola gives,

of
See the title ' The angle between the focal radii at a point of the ellipse'.
 Construction of tangents from a point outside the hyperbola

With A as center draw an arc through F2, and from F1as center, draw an arc of radius 2a.

These arcs intersect at points S1 and S2.

Tangents are the perpendicular bisectors of the line segments F2S1 and F2S2.

Tangents can also be drawn as lines through A and the intersection points of lines through F1S1 and F1S2, with the hyperbola.

These intersections are at the same time the points of contact D1 and D2.

Example:  The line 13x - 15y - 25 = 0 is the tangent of a hyperbola with linear eccentricity (half the focal distance) cH = Ö41.  Write the equation of the hyperbola.

Solution:  Rewrite the equation  13x - 15y - 25 = 0  or
Using the linear eccentricity
and the tangency condition
hus, the equation of the hyperbola,

Example:  Find the normal to the hyperbola 3x2 - 4y2 = 12 which is parallel to the line  - x + y = 0.

Solution:  Rewrite the equation of the hyperbola

      3x2 - 4y2 = 12 | ¸12

The slope of the normal is equal to the slope of the given line,

y = x  =>  m = 1,   mt = -1/mn,  so  mt = -1

applying the tangency condition

a2m2 - b2 = c2 <=  mt = -1, a2 = 4,  b2 = 3
4·(-1)2 - 3 = c2   =>  c1,2 = ±1
tangents  t1 ::  y = -x + 1,   t2 ::  y = -x - 1.
The points of tangency,

The equations of the normals,

D1(4, -3) and  m = 1   =>  y - y1 = m ·(x -x1),       y + 3 = 1·(x - 4)  or   n1 ::   y = x - 7,
D2(-4, 3) and  m = 1  =>  y - y1 = m ·(x -x1),         y - 3 = 1·(x + 4)   or   n2 ::   y = x + 7.
 
 
 
 
 
 
 
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