Hyperbola - 2
FOCAL DISTANCES OF A POINT:
Another Defination of hyperbola is that the difference of focal distances of any point on hyperbola is constant & equal to the length of transverse axis of the hyperbola.
Why ?
Let the hyperbola be
We know that
Distance from focus = (Distance from Directrix)
Hence Distance of point P(x1, y2) from S,(ae,0) is
SP = ePM = e = ex1 - a
Similarly S’P = e(PM') = e = ex1 + a
Hence S’P - SP = 2a
= Transverse axis
Hence Hyperbola is the Locus of a point which moves in a plane such that the difference of its distances from two fixed points i.e. foci is constant.

POINT AND HYPERBOLA
The point (x1, y2) lies outside, on ,or inside the hyperbola


Why ?
Let P = (x1, y2) & Q = (x1 y1)
Draw QL perpendicular to x axis then
QL > PL





Adding


But

Hence



When point lies outside. Similarly we can prove that when point lies on of inside the hyperbola Then

LINE AND A HYPERBOLA
The line y = mx + c will cut the hyperbola

Why ?
Let the line be y = mx + c ............................................. (1)
& the hyperbola

Eliminating y form (1) & (2) we get


This is a quadratic is x, hence
Discriminant = b2 - 4ac
= 4m2c2a4 - 4(a2m2 - b2)(a2)(b2 + c2)
= c2 + b2 - a2m2
Line will cut in two points if D > 0


Line will touch the parabola if D = 0
i.e. c2 + b2 - a2m2 = 0

Line will not be touch or be touch a chord of parabola
if D < 0 i.e. c2 + b2 - a2m2 < 0

ILLUSTRATION : 3.
For what values of K will the line y = 3x + K be a chord to the hyperbola

Ans: For this hyperbola we have
a2 = 9 & b2 = 45
From the equation of given line
m = 3 & c = k
Hence we know that for a line to be a chord to hyperbola c2 > a2m2 - b2
i.e. K2 > 9 x 9 - 45



Hence

EQUATIONS OF TANGENT
(a) POINT FORM:
The equation of tangent to the hyperbola


i.e. T = 0 where T =

Why ?
The equation of hyperbola is:

Differentiating w.r.t.x. we get




Equation of tangent when it passes through (x1y1) is
(y - y1) =


Dividing whole equation by a2b2 we get



But



(b) PARAMETRIC FORM:
The equation of tangent to hyperbola

(a sec



Why ?
We have to paranetric equations of hyperbola as
x = a sec


Differentiating both equations we get dx = a sec





Hence dividing these equations we get,

Hence the equations of tangent is (y - b tan


















SLOPE FORM:
The equations of tangent to the hyperbola



Why ?
Let the line with slope m be
y = mx + c, tangent to

Eliminating y from these two equations we get
(a2m2 - b2)x2 + 2 mca2x + a2(c2 + b2) = 0
This is a quadratic equations in x hence for only one solution D should be zero.




Hence the equation of tangent is:
y = mx

Now, we also know that the equation of tangent at (x1, y1) is

Comparing (1) & (2) as they are the same equation we get




Hence the coordinates of point of contact are
