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Mathematics. Session. Hyperbola Session - 1. Introduction. If S is the focus, ZZ´ is the directrix and P is any point on the hyperbola, then by definition. Question. Illustrative Problem.

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Presentation Transcript
Session

Hyperbola

Session - 1

Introduction

If S is the focus, ZZ´ is the directrix and P is any point on the hyperbola, then by definition

Illustrative Problem

Find the equation of hyperbola whose focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity is 2.

Solution :

Let S(1, 2) be the focus and P(x, y) be any

point on the hyperbola.

where PM = perpendicular distance from P to directrix 3x + 4y + 8 = 0

(ii) Transverse and Conjugate Axes

(i) Vertices

(iii) Foci : As we have discussed earlier S(ae, 0) and S´(–ae, 0) are the foci of the hyperbola.

(iv) Directrices :The lines zk and z´k´

are two directrices of the hyperbola and

their equations are

respectively.

(vi) Eccentricity

For the hyperbola we have

(v) Centre :The middle point O of AA´ bisects every chord of the hyperbola passing through it and is called the centre of the hyperbola.

(ix) Focal Distance of a Point

SP = ex – a

S´P = ex + a

(vii) Ordinate and Double ordinate

(viii) Latus rectum

“A hyperbola is the locus of a point which moves in such a way that the difference of its distances from two fixed points (foci) is always constant.”

If is the equation

Of hyperbola, then its auxiliary circle is x2 + y2 = a2

= eccentric angle

are known as parametric equation

of hyperbola.

The circle drawn on transverse axis of the hyperbola as diameter is called an auxiliary circle of the hyperbola.

Let the equation of line is y = mx + c

and equation of hyperbola is

Intersection of a Line and a Hyperbola

Point of intersection of line and hyperbola could be found out by solving the above two equations simultaneously.

Intersection of a Line and a Hyperbola

[Putting the value of y in the equation of Hyperbola]

This is a quadratic equation in x and therefore gives two values of x which may be real and distinct, coincident or imaginary.

Given hyperbola is

and given line is y = mx + c

Condition for Tangency and Equation of Tangent in Slope Form and Point of contact

This is the required condition for tangency.

Equation of Tangent in Slope Form

Substituting the value of c in the equation y = mx + c, we get equation of tangent in slope form.

Equation of tangent

Point of Contact

Equation of Tangent and Normal in Point Form

Equation of tangent at any point

(x1, y1)of the hyperbola is

Equation of Normal at any point

(x1, y1)of the hyperbola is

Class Exercise - 1

Find the equation to the hyperbola for which eccentricity is 2, one of the focus is (2, 2) and corresponding directrix is x + y – 9 = 0.

According to the definition of hyperbola

Solution

Let P(x, y) be any point of hyperbola.

Let S(2, 2) be the focus.

This is the required equation of hyperbola.

Find the coordinates of centre, lengths of the axes, eccentricity, length of latus rectum, coordinates of foci, vertices and equation of directrices of the hyperbola

Class Exercise - 2

The equation (i) becomes

The coordinates of centre with respect to oldaxes are x – 1 = 0 and y – 2 = 0.

Solution contd..

Shifting the origin at (1, 2) withoutrotating the coordinate axes, i.e.

Put x – 1 = X and y – 2 = Y

Centre: The coordinates of centre with respect to new axes are X = 0 and Y = 0.

x = 1, y = 2

Solution contd..

Length of axes

Length of transverse axes = 2b

Length of conjugate axes = 2a

Eccentricity

Coordinates of foci with respect to old axes are(1, 5) and (1, –1).

Vertices: The coordinates of vertices with respect tonew axes are X = 0 and , i.e. X = 0 and

Solution contd..

Length of latus rectum

The coordinates of axes with respect toold axes are x – 1 = 0, i.e. x = 1 and

Vertices

Directrices: The equation of directrices with respectto new axes are , i.e. .

The equation of directrices with respect toold axesare , i.e. y = 3 and y = 1.

Solution contd..
Class Exercise - 3
• Find the equation of hyperbola whose
• direction of axes are parallel to
• coordinate axes if
• vertices are (–8, –1) and (16, –1) and focus is (17, –1) and
• focus is at (5, 12), vertex at (4, 2) and centre at (3, 2).

(i)Centre of hyperbola is mid-point ofvertices

Equation of hyperbola is

Equation of hyperbola in new coordinate axes is.

Solution

Let x – 4 = X, y + 1 = Y.

Solution contd..

As per definition of hyperbola

a = Distance between centre and vertices

= 144

Abscissae of focus in new coordinates system isX = ae, i.e. x – 4 = 12e

Equation of hyperbola is

(ii) Coordinates of centre are (3, 2).

Equation of hyperbola is

a = Distance between vertex and centre

Equation (i) becomes

Solution contd..

Let x – 3 = X, y – 2 = Y.

Abscissae of focus is X = ae

5 = e + 3 [Abscissae of focus = 5]

= 1 (4 – 1) = 3

Solution contd..

i.e. x – 3 = e (As a = 1)

x = e + 3

Class Exercise - 4

Find the equations of the tangents to the hyperbola 4x2 – 9y2 = 36 which are parallel to the line 5x – 3y = 2.

Any tangent to the hyperbolais

be the middle point of AB

Solution

Let the tangent (i) intersect the x-axis at A and y-axis at B respectively.

Let P(h, k) be the middle point of AB.

Equations (i) and (ii) will represent the same line if

Solution

The equation of the given line is

lx + my + n = 0 ...(ii)

The curve represents

• a hyperbola if k < 8
• an ellipse if k > 8
• a hyperbola if 8 < k < 12
• None of these
Class Exercise - 7

The given equation

represents hyperbola if

(12 – k) (8 – k) < 0

i.e. 8 < k < 12

Solution

Let where be two points on thehyperbola If (h, k) is the point of intersection of the normals at P and Q, then k is equal to

(a) (b)

(c) (d)

Class Exercise - 9

The equation of other hyperbola can be written as

If equations (i) and (ii) are the same, then

Solution