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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.

For more on rectangular hyperbola, view the following video

Some Important Points to be noted:

  • 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.

  • 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 x2 – y2 = ais reduced to xy = a2/2 or xy = c2.

  • When xy = c2, 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 = c2 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(x1, y1) is x/x1 + y/y1 = 2 and at P(t) is x/t + ty = 2c.

  • Equation of normal is y-c/t = t2(x-ct).

  • 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.  

  • 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 12x2 + 9xy + 3x -16xy – 12y2 – 4y + 36x + 27y + 9 – 9 = 0.

So, 12x2 – 12y- 7xy + 39x + 23y = 0.

Related Resources:

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