A hyperbola is defined as an open curve having two branches which are mirror images to each other. Hyperbolic curves are of special importance in astronomy & space studies. Here, we will be studying about hyperbola and characteristics of such curves.
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What is Hyperbola?
A hyperbola is a locus of points in such a way that the distance to each focus is a constant greater than one. In other words, the locus of a point moving in a plane in such a way that the ratio of its distance from a fixed point (focus) to that from a fixed line (directrix) is a constant greater than 1.
[Note: The point (focus) does not lie on the line (directrix)]
(PS/PM) = e > 1 eccentricity
Standard Equation of Hyperbola
The equation of the hyperbola is simplest when the centre of the hyperbola is at the origin and the foci are either on the x-axis or on the y-axis. The standard equation of a hyperbola is given as:
[(x^{2} / a^{2}) – (y^{2} / b^{2})] = 1
where , b^{2} = a^{2} (e^{2} – 1)
Important Terms and Formulas of Hyperbola
There are certain terms related to a hyperbola which needs to be thoroughly understood to be able to get confident with this concept. Some of the most important terms related to hyperbolae are:
- Eccentricity (e): e^{2} = 1 + (b^{2} / a^{2}) = 1 + [(conjugate axis)^{2} / (transverse axis)^{2}]
- Focii: S = (ae, 0) & S′ = (−ae, 0)
- Directrix: x=(a/e), x = (−a / e)
- Transverse axis:
The live segment A’A of length 2a in which the focii S’ and S both lie is called the transverse axis of the hyperbola.
- Conjugant axis:
The line segment B’B of length 2b between the 2 points B’ = (0, -b) & B = (0, b) is called the conjugate axis of the hyperbola.
- Principal axes:
The transverse axis & conjugate axis.
- Vertices:
A = (a, 0) & A’ = (-a, 0)
- Focal chord:
A chord which passes through a focus is called a focal chord.
- Double ordinate:
Chord perpendicular to the transverse axis is called a double ordinate.
- Latus Rectum:
Focal chord ⊥^{r} to the transverse axis is called latus rectum.
Its length = (2b^{2} / a) = [(conjugate)^{2} / transverse] = 2a (e^{2 }− 1)
The difference in focal distances is a constant
i.e. |PS−PS′| = 2a
Length of latus rectum = 2 e × (distance of focus from corresponding directrix)
End points of L.R : (± ae, ± b^{2} / a)
Centre:
The point which bisects every chord of the conic, drawn through it, is called the centre of the conic.
C: (0, 0) is the centre of [(x^{2} / a^{2}) – (y^{2} / b^{2})] = 1
Note:
You will notice that the results for ellipse are also applicable for a hyperbola. You need to replace b^{2} by (−b^{2})
Practice Problems on Hyperbola
Example 1:
Find the equation of the hyperbola whose directrix is 2x + y = 1, focus (1, 2) and eccentricity √3.
Solution:
Let P(x, y) be any point on the hyperbola.
Draw PM perpendicular from P on the directrix,
Then by definition SP=ePM.
⇒ (SP)^{2} = e^{2} (PM)^{2}
⇒ (x − 1)^{2} + (y − 2)^{2} = 3{(2x + y – 1) / √(4+1)}^{2}
⇒ 5 (x^{2} + y^{2} − 2x − 4y + 5)
= 3 (4x^{2} + y^{2} + 1 + 4xy − 2y − 4x)
⇒ 7x^{2} − 2y^{2} + 12xy − 2x + 14y – 22 = 0
Which is the required hyperbola.
Example 2:
Find the eccentricity of the hyperbola whose latus rectum is half of its transverse axis.
Solution:
Let the equation of hyperbola be [(x^{2} / a^{2}) – (y^{2} / b^{2})] = 1
Then transverse axis = 2a and latus – rectum = (2b^{2} / a)
According to question (2b^{2} / a) = (1/2) × 2a
⇒ 2b^{2} = a^{2} (Since, b^{2} = a^{2} (e^{2} − 1))
⇒ 2a^{2} (e^{2} − 1) = a^{2}
⇒ 2e^{2} – 2 = 1
⇒ e^{2} = (3 / 2)
∴ e = √(3/2)
Hence the required eccentricity is √(3/2)
What is Conjugate Hyperbola?
2 hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate & transverse axis of the other are called conjugate hyperbola of each other.
2 hyperbolas such that transverse & conjugate axes of one hyperbola are respectively the conjugate & transverse axis of the other are called conjugate hyperbola of each other.
(x^{2} / a^{2}) – y^{2} b^{2} = 1 & (−x^{2} / a^{2}) + (y^{2} / b^{2}) = 1 are conjugate hyperbolas of each other.
(y^{2} / b^{2}) – (x^{2} / a^{2}) = 1
a^{2 }= b^{2} (e^{2} − 1)
Vertices: Vertices: (0,±b) L.R. = (2a^{2} / b)
Some Important Conclusions on Conjugate Hyperbola
(a) If are eccentricities of the hyperbola & its conjugate, the
(1 / e_{1}^{2}) + (1 / e_{2}^{2}) = 1
(b) The foci of a hyperbola & its conjugate are concyclic & form the vertices of a square.
(c) 2 hyperbolas are similar if they have the same eccentricities.
(d) 2 similar hyperbolas are equal if they have the same latus rectum.
Example 3:
Find the lengths of transverse axis and conjugate axis, eccentricity, the co-ordinates of foci, vertices, length of the latus-rectum and equations of the directrices of the following hyperbola 16x^{2} − 9y^{2} = −144.
Solution:
The equation 16x^{2} − 9y^{2} = −144 can be written as (x^{2} / 9) – (y^{2} / 16) = −1
This is of the form (x^{2} / a^{2}) – (y^{2} / b^{2}) = −1
∴ a^{2} = 9, b^{2} = 16
⇒ a=3, b=4
Length of transverse axis: The length of transverse axis = 2b = 8
Length of conjugate axis: The length of conjugate axis = 2a = 6
Eccentricity:
Foci: The co-ordinates of the foci are (0, ± be) i.e., (0, ± 5)
Vertices: The co-ordinates of the vertices are (0, ± b) i.e. (0, ± 4)
Lengths of latus-rectum: The length of latus-rectum = (2a^{2} / b) = [2(3)^{2}] / 4 = 9/2
Equation of directrices: The equation of directrices are
y = ± (b/e) ⇒y = ± [4 / (5/4)] ⇒ y ± (16 / 5)
Auxiliary Circles of the Hyperbola
A circle drawn with centre C & transverse axis as a diameter is called the auxiliary circle of the hyperbola. The auxilary circle of hyperbola equation is given as:
Equation of the auxiliary circle is x^{2} + y^{2} = a^{2},
Note from the following figure that P & Q are called the “corresponding points” of the hyperbola & the auxiliary circle.
Parametric Representation:
x = a sec θ and y = b tan θ
i.e. If (a sec θ, b tan θ) is on the hyperbola, then Q : (a cos θ, a sin θ) lies on the auxiliary circle. The equation of chord joining 2 points P (α) and Q (f) is given by:
The position of a point P
S_{1} = x_{1}^{2}/a^{2} + y^{2}/b^{2} = 1 is positive, zero or negative accordingly as (x_{1} ,y_{1}) lies inside, on or outside is positive, zero or negative accordingly as (x_{1} ,y_{1}) lies inside, on or outside.
Example 4: Find the position of the point (5, -4) relative to the hyperbola 9x^{2} – y^{2} = 1.
Solution:
Since 9(5)^{2} – (-4)^{2 }– 1 = 225 – 16 – 1 = 208 > 0,
So, the point (5, -4) lies inside the hyperbola 9x^{2} – y^{2} = 1.
Rectangular Hyperbola
The rectangular hyperbola is a hyperbola axes (or asymptotes) are perpendicular, or with its eccentricity is √2. Hyperbola with conjugate axis = transverse axis is a = b example of rectangular hyperbola.
x^{2}/a^{2} – y^{2}/b^{2}
⇒ x^{2}/a^{2 }– y^{2}/a^{2 }= 1
Or, x^{2} – y^{2 }= a^{2}
We know b^{2} = a^{2} (e^{2} − 1) a^{2}
= a^{2} (e^{2} − 1) e^{2} = 2e = √2.
Eccentricity of rectangular hyperbola
Also, xy = c
e.g. xy = 1, y = 1/x
Tangent of Rectangular hyperbola
The tangent of a rectangular hyperbola is a line that touches a point on the rectangular hyperbola’s curve. The equation and slope form of a rectangular hyperbola’s tangent is given as:
Equation of tangent
The y = mx + c write hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1 will be tangent if c^{2} = a^{2}/m^{2} – b^{2}.
Slope form of tangent
y = mx ± √(a^{2}m^{2} – b^{2})
Secant
Secant will cut ellipse at 2 distinct points
⇒ c^{2} > a^{2} m^{2} – b^{2}
Neither Secant Nor Tangent
For line to be neither secant nor tangent, quadratic equation will give imaginary solution.
⇒ c^{2} < a^{2} m^{2} – b^{2}
Equation of tangent to hyperbola x^{2} /a^{2} – y^{2}/b^{2} = 1 at point (x_{1} ,y_{1})
= (xx_{1})/a2 = (yy_{1})/b_{1} = 1
Parametric form of tangent:
(x secθ) /a − (y tanθ)/b = 1
Point of contact and examples on tangent
Compare:
y = mx + c
(xx_{1})/a^{2} – (yy_{1})/b^{2} = 1 – mx + y = c
x_{1} = (-a^{2}c)/m;
y_{1} = -b^{2}/c
(x_{1},y_{1}) = [(-a^{2}m)/c, -b^{2}/c]
Solved Examples on Hyperbola
Example 5: Show that the line x cos α + y sin α = p touches the hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1. If a^{2} cos^{2} α – b^{2} sin^{2} α = p^{2}.
Solution:
Given line is x cos α + y sinα = p
y = − x cot α + p cosec α
Condition for y = mx + c to be tangent of hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1.
c^{2} = a^{2}/m^{2} – b^{2}
p^{2} cosec2α = α^{2} cot2α – b^{2}
p^{2} = a^{2} cos^{2}α – b^{2} sin^{2} α
Director circle:
Locus of point of intersection of ⊥r tangents.
Director circle for x^{2}/a^{2} – y^{2} /b^{2} = 1 is
x^{2} + y^{2} = a^{2 }– b^{2}
Example 6: Find the equation of tangent to hyperbola x29−y2=1 whose slope is 5
Solution:
Slope of tangent m = 5, a^{2} = 9, b^{2} = 1
Equation of tangent in slope form is
y = mx ± √(a^{2}m^{2} – b^{2} )
y = 5x ± √(9.5^{2 }– 1)
y = 5x ± 4√14
[Note: For ellipse, director circle is x^{2} + y^{2} = a^{2} + b^{2}, x^{2}/a^{2} + y^{2}/b^{2} = 1]
Normal:
Equation of normal of x^{2}/a^{2} – y^{2}/b^{2} = 1 at (x_{1} ,y_{1})
a^{2}x/x_{1} + b^{2}y/y_{1} = a^{2} + b^{2}
Normal in parametric form:
ax/secθ + by/tanθ = a^{2} + b^{2}
Example 7: Find normal at the point (6, 3) to hyperbola x^{2}/18 − y^{2}/9 = 1
Solution:
Equation of Normal at point (x_{1}, y_{1}) is a^{2 }= 18, b^{2} = 9
a^{2}x/x_{1} + b^{2}y/y_{1} = a^{2} + b^{2}
Equation of Normal at point (6, 3) is
18x/6 + 9y/3 = 18 + 9
x + y = 9
Chord of contact:
T = 0
xx1/a^{2} − yy1/b^{2} = 1
Example 8: Find equation of chord of Contact of point (2, 3) to hyperbola x^{2}/16 − y^{2}/9 = 1
Solution:
Equation of chord of Contact is T = 0
i.e. (xx_{1})/a^{2} – (yy_{1})/b^{2} – 1 = 0
Or, 2x/16 – 3y/9 = 1
Or, x/8 – y/3 = 1
Equation of chord when mid-point is given
T = (xx_{1})/a^{2} – (yy_{1})/b^{2} – 1 = x_{1}^{2}/a^{2} – y_{1}^{2}/b^{2} − 1.
Example 9: Find the equation of chord of hyperbola x^{2}/9 – y^{2}/4 = 1 whose midpoint is (5, 1).
Solution:
Equation of chord of the hyperbola whose mid points is (5, 1)
T = (xx_{1})/a^{2} – (yy_{1})/b^{2} – 1 = x_{1}^{2}/a^{2} – y_{1}^{2}/b^{2} − 1
5x/9 – y/4 – 1= 25/9 – ¼ – 1
5x/9 – y/4 = 91/36