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The hyperbola whose asymptotes are at right angles to each other is called a rectangular hyperbola. The angle between asymptotes of the hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 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 x^{2} – y^{2} = a^{2}.
We know that the asymptotes of the hyperbola x^{2}/a^{2} – y^{2}/b^{2} = 1 …… (1)
are given by y = + (b/a) x …… (2)
If θ be the angle between the asymptotes, then
θ = tan^{–1} ((m_{1}–m_{2})/(1 + m_{1}m_{2}))
= tan^{–1} [{(b/a)–(–b/a)}/{1+(b/a)(–b/a)}]
= tan^{–1} [2(b/a)/(1–(b^{2}/a^{2}))]
= 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 x^{2} – y^{2} = a^{2}.
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 -45^{o}.
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} = a^{2} i.e.
xy = ½ a^{2} or xy = c^{2} where c^{2} = a^{2}/2.
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In a hyperbola b^{2} = a^{2} (e^{2} – 1). In the case of rectangular hyperbola (i.e., when b = a) result becomes a^{2} = a^{2}(e^{2} – 1) or e^{2} = 2 or e = √2
i.e. the eccentricity of a rectangular hyperbola = √2.
In case of rectangular hyperbola a = b i.e., the length of transverse axis = length of conjugate axis.
A rectangular hyperbola is also known as an equilateral hyperbola.
The asymptotes of rectangular hyperbola are y = ± x.
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 x^{2} – y^{2} = a^{2 }is reduced to xy = a^{2}/2 or xy = c^{2}.
When xy = c^{2}, the asymptotes are the coordinate axis.
Length of latus rectum of rectangular hyperbola is the same as the transverse or conjugate axis.
Rectangular Hyperbola with asymptotes as coordinate axis:
The equation of the hyperbola which has its asymptotes as the coordinate axis is xy = c^{2} with parametric representation x = ct and y = c/t, t ∈ R-{0}.
The equations of the directrices of the hyperbola in this case are x + y = ± √2c.
Since, the transverse and the conjugate axes are the same hence, length of latus rectum = 2√2c = T.A. = C.A.
Equation of a chord whose middle point is given to be (p, q) is qx + py = 2pq.
The equation of the tangent at the point P(x_{1}, y_{1}) is x/x_{1} + y/y_{1} = 2 and at P(t) is x/t + ty = 2c.
Equation of normal is y-c/t = t^{2}(x-ct).
The equation of the chord joining the points P(t_{1}) and Q(t_{2}) is x + t_{1}t_{2}y = c(t_{1} + t_{2}) and its slope is m = -1/t_{1}t_{2}.
The vertices of the hyperbola are (c, c) and (-c, -c) and the focus is (√2c, √2c) and (-√2c, -√2c).
A rectangular hyperbola circumscribing a triangle passes through the orthocentre of this triangle.
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.
Find the equation of the hyperbola with asymptotes 3x – 4y + 9 = 0 and 4x + 3y + 1 = 0 which passes through the origin.
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 12x^{2} + 9xy + 3x -16xy – 12y^{2} – 4y + 36x + 27y + 9 – 9 = 0.
So, 12x^{2} – 12y^{2 }- 7xy + 39x + 23y = 0.
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