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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Session - 1
If S is the focus, ZZ´ is the directrix and P is any point on the hyperbola, then by definition
Find the equation of hyperbola whose focus is (1, 1), directrix is 3x + 4y + 8 = 0 and eccentricity is 2.
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
(iii) Foci : As we have discussed earlier S(ae, 0) and S´(–ae, 0) are the foci of the hyperbola.
(iv) Directrices Hyperbola :The lines zk and z´k´
are two directrices of the hyperbola and
their equations are
For the hyperbola we haveDefinition of Special Points Lines of the Equation of Hyperbola
(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 Hyperbola
SP = ex – a
S´P = ex + aDefinition of Special Points Lines of the Equation of Hyperbola
(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.”
The conjugate hyperbola
of the hyperbola
If Hyperbola is the equation
Of hyperbola, then its auxiliary circle is x2 + y2 = a2
= eccentric angle
are known as parametric equation
of hyperbola.Auxiliary Circle and Eccentric AngleParametric Coordinate of Hyperbola
The circle drawn on transverse axis of the hyperbola as diameter is called an auxiliary circle of the hyperbola.
insidePosition of Point with respect to Hyperbola
Let the equation of line is y = mx + c Hyperbola
and equation of hyperbola isIntersection of a Line and a Hyperbola
Point of intersection of line and hyperbola could be found out by solving the above two equations simultaneously.
[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 Hyperbola
and given line is y = mx + cCondition for Tangency and Equation of Tangent in Slope Form and Point of contact
This is the required condition for tangency.
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 at any point
(x1, y1)of the hyperbola is
Equation of Normal at any point
(x1, y1)of the hyperbola is
Equation of tangent at Hyperbola
isEquation of Tangent and Normal in Parametric Form
Equation of normal in parametric form is
Class Test Hyperbola
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 HyperbolaSolution
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 hyperbolaClass Exercise - 2
The given equation can be written as eccentricity, length ofSolution
The equation (i) becomes eccentricity, length of
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 eccentricity, length ofSolution contd..
Length of axes
Length of transverse axes = 2b
Length of conjugate axes = 2a
Foci: eccentricity, length of Coordinates of foci with respect to new axes are, i.e..
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 andSolution contd..
Length of latus rectum
The coordinates of axes with respect to eccentricity, length of old axes are x – 1 = 0, i.e. x = 1 and
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..
(i) eccentricity, length of 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.
As per definition of hyperbola
a = Distance between centre and vertices
Abscissae of focus in new coordinates system isX = ae, i.e. x – 4 = 12e
Equation of hyperbola is eccentricity, length of
(ii) Coordinates of centre are (3, 2).
Equation of hyperbola is
a = Distance between vertex and centre
Equation (i) becomesSolution contd..
Let x – 3 = X, y – 2 = Y.
Abscissae of focus is X = ae eccentricity, length of
5 = e + 3 [Abscissae of focus = 5]
= 1 (4 – 1) = 3Solution contd..
i.e. x – 3 = e (As a = 1)
x = e + 3
Find the equations of the tangents to the hyperbola 4x2 – 9y2 = 36 which are parallel to the line 5x – 3y = 2.
Tangent is parallel to the given line 5x – 3y = 2 eccentricity, length of
Equation of tangentsSolution
Find the locus of mid-point of portion of tangent intercepted between the axes for hyperbolaClass Exercise - 5
Any tangent to the hyperbola intercepted between the axes is
be the middle point of ABSolution
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.
Find the condition that the line intercepted between the axes lx + my + n = 0 will be normal to the hyperbolaClass Exercise - 6
Equations (i) and (ii) will represent the same line if intercepted between the axesSolution
The equation of the given line is
lx + my + n = 0 ...(ii)
The given equation intercepted between the axes
represents hyperbola if
(12 – k) (8 – k) < 0
i.e. 8 < k < 12Solution
Hence, answer is (c).
If the line intercepted between the axes touches the hyperbola at the point , show thatClass Exercise - 8
Both (i) and (ii) represent same line intercepted between the axesSolution contd..
Let intercepted between the axes 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
(c) (d)Class Exercise - 9
From (iv) and (v) eliminating h, we get intercepted between the axesSolution contd..
Determine the equations of common tangent to the hyperbola intercepted between the axesClass Exercise - 10
The equation of other hyperbola can be written as intercepted between the axes
If equations (i) and (ii) are the same, thenSolution
Thank you intercepted between the axes