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^–1b/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 -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 = a² i.e.

xy = ½ a² or xy = c² where c² = a²/2.

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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 = a²is 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.

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.

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