how the rectangular hyperbola equation :xy=c^2 comes

We realize that the asymptotes of the hyperbola x2/a2 – y2/b2 = 1 … (1) 

are given by y = + (b/a) x … (2) 

On the off chance that θ be the edge 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) 

Be that as it may, if the hyperbola is rectangular, then θ = π/2 

i.e., π/2 = 2 tan–1 (b/an) or tan (π/4) = b/a 

⇒ b = a 

Consequently, from (1) the condition of the rectangular hyperbola is x2 – y2 = a2. 

With a specific end goal to get the condition of the hyperbola which has asymptotes as facilitate hub we pivot the tomahawks of reference through an edge of - 45o. 

Subsequently, for this we need to compose x/√2 + y/√2 for x and –x/√2 + y/√2 for y. 

The condition (i) progresses toward becoming 

(1/2)(x + y)2 – (1/2)(x – y)2 = a2 i.e. 

xy = ½ a2 or xy = c2 where c2 = a2/2.