Question

Why is it called a rectangular hyperbola?

Answer 1

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 ${{a}^{2}}={{a}^{2}}\left( {{e}^{2}}-1 \right)$ 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 $y-\dfrac{c}{t}={{t}^{2}}\left( x-ct \right)$. 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, $\left( 3x-4y+9 \right)\left( 4x+3y+1 \right)+r=0$
Since we are told that the hyperbola will be passing through the origin, we will be substituting $x=y=0$.
Upon substituting we get,

& 9+r=0 \\\ & \Rightarrow r=-9 \\\ \end{aligned}$$ Now, we obtain the equation of hyperbola as $$\begin{aligned} & \left( 3x-4y+9 \right)\left( 4x+3y+1 \right)-9=0 \\\ & \Rightarrow 12{{x}^{2}}+9xy+3x-16xy-12{{y}^{2}}-4y+36x+27y+9-9=0 \\\ & \Rightarrow 12{{x}^{2}}-12{{y}^{2}}+39x+23y-7xy=0 \\\ \end{aligned}$$  **Note:** e must always remember that the eccentricity of the rectangular hyperbola is $$\sqrt{2}$$. We must also note that the length of latus rectum can be calculated by the formula $$2\sqrt{2}c$$ and when the vertices of a hyperbola are $$\left( c,c \right)$$ and $$\left( -c,-c \right)$$, then the focus of the hyperbola is $$\left( \sqrt{2}c,\sqrt{2}c \right)$$ and $$\left( -\sqrt{2}c,-\sqrt{2}c \right)$$.