Rectangular Hyperbola
What do we mean by a Rectangular Hyperbola?
The hyperbola whose asymptotes are at right angles to each other is called a rectangular hyperbola. The angle between asymptotes of the hyperbola x2/a2 – y2/b2 = 1, is 2 tan–1 (b/a).
This is a right angle if tan–1 b/a = π/4, i.e., if b/a = 1 ⇒ b = a.
The equation of rectangular hyperbola referred to its transverse and conjugate axes as axes of coordinates is therefore x2 – y2 = a2.
How do we compute the Rectangular Hyperbola Equation?
We know that the asymptotes of the hyperbola x2/a2 – y2/b2 = 1 …… (1)
are given by y = + (b/a) x …… (2)
If θ be the angle between the asymptotes, then
θ = tan–1 ((m1–m2)/(1 + m1m2))
= tan–1 [{(b/a)–(–b/a)}/{1+(b/a)(–b/a)}]
= tan–1 [2(b/a)/(1–(b2/a2))]
= 2 tan–1 (b/a)
But if the hyperbola is rectangular, then θ = π/2
i.e., π/2 = 2 tan–1 (b/a) or tan (π/4) = b/a
⇒ b = a
Therefore, from (1) the equation of the rectangular hyperbola is x2 – y2 = a2.
In order to obtain the equation of the hyperbola which has asymptotes as coordinate axis we rotate the axes of reference through an angle of -45o.
Hence, for this we have to write x/√2 + y/√2 for x and –x/√2 + y/√2 for y.
The equation (i) becomes
(1/2)(x + y)2 – (1/2)(x – y)2 = a2 i.e.
xy = ½ a2 or xy = c2 where c2 = a2/2.
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Some Important Points to be noted:
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In a hyperbola b2 = a2 (e2 – 1). In the case of rectangular hyperbola (i.e., when b = a) result becomes a2 = a2(e2 – 1) or e2 = 2 or e = √2
i.e. the eccentricity of a rectangular hyperbola = √2.
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In case of rectangular hyperbola a = b i.e., the length of transverse axis = length of conjugate axis.
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A rectangular hyperbola is also known as an equilateral hyperbola.
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The asymptotes of rectangular hyperbola are y = ± x.
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If the axes of the hyperbola are rotated by an angle of -π/4 about the same origin, then the equation of the rectangular hyperbola x2 – y2 = a2 is reduced to xy = a2/2 or xy = c2.
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When xy = c2, the asymptotes are the coordinate axis.
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Length of latus rectum of rectangular hyperbola is the same as the transverse or conjugate axis.
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Rectangular Hyperbola with asymptotes as coordinate axis:
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The equation of the hyperbola which has its asymptotes as the coordinate axis is xy = c2 with parametric representation x = ct and y = c/t, t ∈ R-{0}.
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The equations of the directrices of the hyperbola in this case are x + y = ± √2c.
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Since, the transverse and the conjugate axes are the same hence, length of latus rectum = 2√2c = T.A. = C.A.
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Equation of a chord whose middle point is given to be (p, q) is qx + py = 2pq.
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The equation of the tangent at the point P(x1, y1) is x/x1 + y/y1 = 2 and at P(t) is x/t + ty = 2c.
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Equation of normal is y-c/t = t2(x-ct).
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The equation of the chord joining the points P(t1) and Q(t2) is x + t1t2y = c(t1 + t2) and its slope is m = -1/t1t2.
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The vertices of the hyperbola are (c, c) and (-c, -c) and the focus is (√2c, √2c) and (-√2c, -√2c).
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A rectangular hyperbola circumscribing a triangle passes through the orthocentre of this triangle.
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If a circle intersects a rectangular hyperbola at four points, then the mean value of the points of intersection is the mid-point of the line joining the centres of both circle and hyperbola.
Illustration:
Find the equation of the hyperbola with asymptotes 3x – 4y + 9 = 0 and 4x + 3y + 1 = 0 which passes through the origin.
Solution:
The asymptotes are given as 3x – 4y + 7 = 0 and 4x + 3y + 1 = 0.
Hence, the joint equation of the asymptotes is (3x – 4y + 9)(4x + 3y + 1) = 0.
Now, we know that the equation of rectangular hyperbola differs from that of the joint equation of asymptotes by a constant only, hence we must have the equation of the hyperbola as
(3x – 4y + 9)(4x + 3y + 1) + r = 0.
It is given that the hyperbola passes through the origin and hence, putting x = y = 0 , we obtain,
9 + r = 0
Hence, r = -9.
Hence, the equation of the hyperbola is (3x – 4y + 9)(4x + 3y + 1) – 9 = 0.
This gives 12x2 + 9xy + 3x -16xy – 12y2 – 4y + 36x + 27y + 9 – 9 = 0.
So, 12x2 – 12y2 - 7xy + 39x + 23y = 0.
Related Resources
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