Question

Why is it called a rectangular hyperbola?

Answer 1

Hint: For a hyperbola to be called a rectangular hyperbola, we must check for the eccentricity of the hyperbola. If the eccentricity of the hyperbola is $$\sqrt{2}$$, then the hyperbola is considered as a rectangular hyperbola. The equation of the rectangular hyperbola is $$a^2=a^2(e^2-1)$$ where $$a=b$$ for a rectangular hyperbola.

Complete step-by-step solution:
Now let us learn more about rectangular hyperbola. In a rectangular hyperbola, $$a=b$$ i.e. the length of the transverse axis = length of the conjugate axis. The asymptotes of the rectangular hyperbola are $$y=\pm x$$. When $$xy=c^2$$, the asymptotes are the coordinate axes. The equation of a normal rectangular hyperbola is $$y-{\displaystyle \frac{c}{t}}=t^2(x-ct)$$. A hyperbola for which the asymptotes are perpendicular, it is called an equilateral hyperbola or right hyperbola. The main difference between a regular hyperbola and rectangular hyperbola is that the asymptotes are perpendicular in the rectangular hyperbola. The equation of a rectangular hyperbola is $$x^2-y^2=a^2$$. The AFC curve is represented by a rectangular hyperbola.
Now let us see why a hyperbola is called a rectangular hyperbola.
When a hyperbola has its asymptotes or the axes perpendicular to each other then it is called a rectangular hyperbola. Also, the eccentricity of a rectangular hyperbola is $$\sqrt{2}$$.
Now, let us find the equation of the rectangular hyperbola whose asymptotes are $$3x-4y+9=0$$ and $$4x+3y+1=0$$ which passes through the origin.
We know that the join equation of the asymptotes and the equation of the hyperbola differs only by a constant $$r$$.
So we get, $$(3x-4y+9)(4x+3y+1)+r=0$$
Since we are told that the hyperbola will be passing through the origin, we will be substituting $$x=y=0$$.
Upon substituting we get,
$$\begin{array}{rl}& 9+r=0\\ & \Rightarrow r=-9\end{array}$$
Now, we obtain the equation of hyperbola as
$$\begin{array}{rl}& (3x-4y+9)(4x+3y+1)-9=0\\ & \Rightarrow 12x^2+9xy+3x-16xy-12y^2-4y+36x+27y+9-9=0\\ & \Rightarrow 12x^2-12y^2+39x+23y-7xy=0\end{array}$$

Note: e must always remember that the eccentricity of the rectangular hyperbola is $$\sqrt{2}$$. We must also note that the length of latus rectum can be calculated by the formula $$2\sqrt{2}c$$ and when the vertices of a hyperbola are $$(c,c)$$ and $$(-c,-c)$$, then the focus of the hyperbola is $$(\sqrt{2}c,\sqrt{2}c)$$ and $$(-\sqrt{2}c,-\sqrt{2}c)$$.