Rectangular Hyperbola: Equation, Graph, Questions, Examples

In coordinate geometry, rectangular hyperbola is a type of hyperbola in which the asymptotes intersect each other at $90^\circ$. In the below article, we will learn more about the rectangular hyperbola, its properties, equation, asymptotes and other characteristics.

This topic falls under the category of coordinate geometry, and is an important chapter in the syllabus of Class 11th mathematics. It is important for both board exams as well as competitive exams such as the JEE Main exam, WBJEE, BITSAT, etc. In total, there are 30 questions which have been asked in the JEE Mains exam in pass 10 years from this topic.

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This Story also Contains

1. What is Rectangular Hyperbola?
2. Equation of Rectangular Hyperbola
3. Rectangular Hyperbola Graph
4. Properties of Rectangular Hyperbola
5. Solved Examples Based on Rectangular Hyperbola

What is Rectangular Hyperbola?

A rectangular hyperbola is a special type of hyperbola whose asymptotes are perpendicular to each other. And the length of the conjugate axis is equal to transverse axis. It is a hyperbola that has transverse axis and conjugate axis of equal length. Its arcs resembles that of a circle.

For a rectangular hyperbola having the transverse axis of length $2 p$ and the conjugate axis of length $2 q$, we have $2 p=$ $2 q$, or $p=q$. The general equation of a rectangular hyperbola is $x^2-y^2=p^2$.

Rectangular Hyperbola Condition

Rectangular hyperbola shape can be imagined as consisting of two curves or branches located in the opposite quadrants (such as first and third quadrant). These branches never touch the asymptotes (x-axis, y-axis).

Hyperbola is symmetric with branches, across the origin. The nature of hyperbola is it is infinite curve, having no intersection with the axis.

Equation of Rectangular Hyperbola

Rectangular hyperbola equation can be denoted using various forms, as per the orientation and center details provided. Below are the equation of rectangular hyperbola:

Standard Equation (centered at the origin)

For a rectangular hyperbola, having asymptotes along the coordinate axis, the standard equation is of the form, $xy = c^2$.

Here $c$ is the constant which shows the size of hyperbola. It has asymptotes along the x axis, y axis.

General Equation (symmetry about the origin)

For the hyperbola, which is symmetric about the origin, the general equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Parametric Form

The parametric equations for the rectangular parabola is $x = ct, \, y = \frac{c}{t}$

Rectangular Hyperbola Graph

A rectangular hyperbola is a type of hyperbola that is specifically defined as having the property that the asymptotes are perpendicular to each other, forming a right angle. Graph of Rectangular Hyperbola with equation $x y=c^2$ where $c$ is a constant that determines the scale of the hyperbola.

Rectangular Hyperbola Shape: If we rotate the coordinate axes by $45^{\circ}$ keeping the origin fixed, then the axes coincide with lines $y$ $=x$ and $y=-x$

Using rotation, the equation $x^2-y^2=a^2$ reduces to
$ \begin{aligned} & xy = \frac{a^2}{2} \\ & \Rightarrow xy = c^2 \end{aligned} $

For rectangular hyperbola, $x y=c^2$

1. Vertices: $\mathrm A(c, c)$ and $\mathrm A^{\prime}(-c,-c)$
2. Transverse axis: $x=y$
3. Conjugate axis: $x=-y$
4. Foci: $\mathrm{S}(c \sqrt{2}, c \sqrt{2})$ and $\mathrm{S}^{\prime}(-c \sqrt{2},-c \sqrt{2})$
5. Directrices: $x+y=\sqrt{ } 2, x+y=-\sqrt{ } 2$
6. Length of latus rectum $=\mathrm{AA}^{\prime}=2 \sqrt{2} c$

Eccentricity of Rectangular Hyperbola

The equation of the rectangular hyperbola is $x^2-y^2=a^2$. Now we know that the eccentricity of the hyperbola is,
$ e = \sqrt{1 + \frac{b^2}{a^2}} $

Rectangular hyperbola eccentricity:
$ \begin{aligned} & a = b \Rightarrow b^2 = a^2 \\ & \Rightarrow a^2 \left( e^2 - 1 \right) = a^2 \\ & \Rightarrow e = \sqrt{2} \end{aligned} $

Asymptotes of a Rectangular Hyperbola

Asymptotes are the lines that connect the curve at infinity. Asymptotes of a Rectangular Hyperbola are Perpendicular. In the case of rectangular hyperbola, the equation of asymptote is,
$ \begin{aligned} & y = \pm x \\ & x^2 - y^2 = 0 \end{aligned} $

Properties of Rectangular Hyperbola

The properties of rectangular hyperbola are,

(i) The parametric equation of the rectangular hyperbola $\mathrm{xy}=\mathrm{c}^2$ are $x=$ ct and $y=\frac{c}{t}$.

(ii) The equation of the tangent to the rectangular hyperbola $x y=c^2$ at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $\mathrm{xy}_1+\mathrm{x}_1 \mathrm{y}=2 \mathrm{c}^2$.

(iii) The equation of the tangent at $\left(\mathrm{ct}, \frac{\mathrm{c}}{\mathrm{t}}\right)$ to the hyperbola $\mathrm{xy}=\mathrm{c}^2$ is $\frac{\mathrm{x}}{\mathrm{t}}+\mathrm{yt}=2 \mathrm{c}$.

(iv) The equation of the normal at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to the hyperbola $\mathrm{xy}=\mathrm{c}^2$ is $\mathrm{xx}_1-\mathrm{yy}_1=\mathrm{x}_1^2-\mathrm{y}_1^2$.

(v) The equation of the normal at $t$ to the hyperbola $x y=c^2$ is $\mathrm{xt}^3-\mathrm{yt}-\mathrm{ct}{ }^4+\mathrm{c}=0$.

(vi) A rectangular hyperbola is symmetric about both its axes and its asymptotes.

Solved Examples Based on Rectangular Hyperbola

Example 1: If the equation $4 \mathrm{x}^2+\mathrm{ky}^2=18$ represents a rectangular hyperbola, then k is equal to
1) 4
2) -4
3) 3
4) None of these

Solution:
Clearly for $4 \mathrm{x}^2+\mathrm{ky}^2=18$ to represent a rectangular hyperbola $\mathrm{k}=4$
Hence, the answer is the option 1.

Example 2: At the point of intersection of the rectangular hyperbola $\mathrm{xy}=\mathrm{c}^2$ and the parabola $\mathrm{y}^2=4 \mathrm{ax}$. The tangents to the rectangular hyperbola and the parabola make an angle $\theta$ and $\phi$ respectively with the axis of X, then

1) $\theta=\tan ^{-1}(-2 \tan \phi)$
2) $\phi=\tan ^{-1}(-2 \tan \theta)$
3) $\theta=\frac{1}{2} \tan ^{-1}(-\tan \phi)$
4) $\phi=\frac{1}{2} \tan ^{-1}(-\tan \theta)$

Solution:

Let $\left(x_1, y_1\right)$ be the point of intersection $\Rightarrow y_1^2=4 a x_1$ and $x_1 y_1=c^2$

For Parabola we have,
$ \begin{aligned} & y^2 = 4ax \\ & \therefore \frac{dy}{dx} = \frac{2a}{y} \\ & \frac{dy}{dx}_{(x_1, y_1)} = \tan \phi = \frac{2a}{y_1} \end{aligned} $

For rectangular hyperbola we have,
$ \begin{aligned} & xy = c^2 \\ & \frac{dy}{dx} = -\frac{y}{x} \\ & \frac{dy}{dx}(x_1, y_1) = \tan \phi = \frac{y_1}{x_1} \\ & \therefore \frac{\tan \theta}{\tan \phi} = \frac{-y_1 / x_1}{2a / y_1} = \frac{-y_1^2}{2a x_1} = -\frac{4a_1}{2a_1} = -2 \\ & \Rightarrow \theta = \tan^{-1}(-2 \tan \phi) \end{aligned} $

Hence, the answer is option 1.

Example 3: Find the foci of the rectangular hyperbola whose equation is $x^2-y^2=16$.
Solution:
Equation of Rectangular Hyperbola is, $x^2-y^2=a^2 \ldots$ (i)
Given Equation,
$x^2-y^2=16$.
$x^2 - y^2 = 4^2$

Comparing Equation (i) and (ii)
$a=4$

Foci of Rectangular Hyperbola is $( \pm a \sqrt{ 2},0$ )
So, Foci of Given Rectangular Hyperbola is $( \pm 4 \sqrt{2},0)$

Example 4: If tangents $OQ$ and $OR$ from $O$ are drawn to a variable circle having radius r and the centre lying on the rectangular hyperbola $x y=1$, then the locus of circumcentre of $\triangle O Q R$ is equal to... ( $O$ is the origin)

Solution:

Let $S\left(t, \frac{1}{t}\right)$ be any point on the given rectangular hyperbola $x y=1$.

A circle is drawn with a centre at $S$ and radius $r$. From origin $O$ tangents $O Q$ and $O R$ are drawn to the above circle. $O Q S R$ is a cyclic quadrilateral.
Hence, points $O, Q, S$ and $R$ are concyclic.

The Circumcircle of $\triangle O Q R$ also passes through $S$ and $O S$ is the diameter.
Therefore, the circumcentre of $\triangle O Q R$ is the mid-point of $O S$. If $(x, y)$ is the circumcentre of $\triangle O Q R$, then

$x=\frac{0+t}{2}, y=\frac{0+\frac{1}{t}}{2}$

$\therefore x y=\frac{1}{4}$

So, the required locus is $x y=\frac{1}{4}$.
Hence, the answer is $x y=\frac{1}{4}$

Example 5: Consider the set of hyperbolas $x y=k, x \in R$. Let $e_1$ be the eccentricity when $k=4$ and $e_2$ be the eccentricity when $k=9$, then $e_1-e_2$ is equal to:
Solution:
We know that the eccentricity of $x y=k$ for all $k \in R$ is $\sqrt{2}$.
$\therefore \quad e_1=\sqrt{2}$ and also $e_2=\sqrt{2}$
Hence,
$e_1-e_2=0$

Hence, the required answer is 0.

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