Hyperbola: Meaning, Equation, Formula, Graph, Shape

The four curves - circle, parabola, ellipse, and hyperbola are called conic sections because they can be formed by interesting a double right circular cone with a plane. The locus of a point moves in a plane such that the ratio of the distance from a fixed point (focus) to the distance from a fixed line (directrix) is constant, the value of which is more than 1. In real life, we use hyperbola to predict the path of the satellite.

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In this article, we will cover the concept of Hyperbola. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of eighteen questions have been asked on JEE MAINS( 2013 to 2023) from this topic including three in 2019, three in 2020, one in 2021, seven in 2022, and one in 2023.

What is a Hyperbola?

A Hyperbola is the set of all points ( $x,y$ ) in a plane such that the difference of their distances from two fixed points is a constant. Each fixed point is called a focus.

Or

The locus of a point moves in a plane such that the ratio of the distance from a fixed point (focus) to the distance from a fixed line (directrix) is constant. The constant is known as eccentricity $e$ and for hyperbola $\mathrm{e}>1$.

Standard Equation Of Hyperbola

The standard form of the hyperbola equation with centre $(0,0)$ and hyperbola foci lying on the $x$-axis is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

where, $b^2=a^2(e^2-1)$

Derivation of Hyperbola Equation

Consider the figure, $O$ is the origin, $S$ and $S^{\prime }$ are the foci and $ZM$ and $Z^{\prime }{M}^{\prime }$ are the directrices.
The foci are $S(ae,0)$ and $S^{\prime }(-ae,0)$. The equation of directrices: $ZM$ is $x=a/e$ and $Z^{\prime }{M}^{\prime }$ is $x=-a/e$ $P(x,y)$ is any point on the hyperbola and $PM$ is perpendicular to directrix $ZM$.

$\begin{array}{rl}& \frac{PS}{PM}=e\Rightarrow (PS{)}^2=e^2(PM{)}^2\\ & (x-ae{)}^2+(y-0{)}^2=e^2{(x-\frac{a}{e})}^2\\ & x^2+a^2{e}^2-2aex+y^2=e^2{x}^2-2aex+a^2\\ & x^2(e^2-1)-y^2=a^2(e^2-1)\\ & \frac{x^2}{a^2}-\frac{y^2}{a^2(e^2-1)}=1\\ & \frac{x^2}{a^2}-\frac{y^2}{b^2}=1,b^2=a^2(e^2-1)\end{array}$

Important Terms related to Hyperbola

• Centre: All chord passing through point $O$ is bisected at point $O$. Here $O$ is the origin, i.e. $(0,0)$.
• Hyperbola Foci: Point $S$ and $S^{\prime }$ are foci of the hyperbola where, $S$ is $(\mathrm{ae},0)$ and $S^{\prime }$ is $(-\mathrm{ae},0)$.
• Directrices: The straight line $ZM$ and $Z^{\prime }{M}^{\prime }$ are two directrices of the hyperbola and their equations are $x=$ ae and $x=-ae$.
• Double Ordinate: If a line perpendicular to the transverse axis of the hyperbola meets the curve at $Q$ and $Q^{\prime }$, then $Q{Q}^{\prime }$ is called double ordinate.
• Latus rectum: Double ordinate passing through focus is called latus rectum. Here $L{L}^{\prime }$ and $L_1{L}_1{}^{\prime }$ are two latus rectum of a hyperbola.
• Vertices: The points where the hyperbola intersects the axis are called the vertices. The vertices of the hyperbola are $(a,0),(-a,0)$.
• Transverse Axis: The line passing through the two foci and the centre of the hyperbola is called the transverse axis of the hyperbola.
• Conjugate Axis: The line passing through the centre of the hyperbola and perpendicular to the transverse axis is called the conjugate axis of the hyperbola.
• Eccentricity of Hyperbola: $(\mathrm{e}>1)$ The eccentricity is the ratio of the distance of the focus from the centre of the hyperbola, and the distance of the vertex from the centre of the

Eccentricity of Hyperbola

The eccentricity is the ratio of the distance of the focus from the center of the hyperbola, and the distance of the vertex from the center of the hyperbola.

Eccentricity of Hyperbola Formula

Equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ we have,

$\begin{array}{rl}& {\mathrm{b}}^2={\mathrm{a}}^2({\mathrm{e}}^2-1)\\ & {\mathrm{e}}^2=\frac{{\mathrm{b}}^2+{\mathrm{a}}^2}{{\mathrm{a}}^2}\\ & \mathrm{e}=\sqrt{1+\left(\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\right)}\\ & \mathrm{e}=\sqrt{1+{\left(\frac{2\mathrm{b}}{2\mathrm{a}}\right)}^2}\\ & \mathrm{e}=\sqrt{1+{\left(\frac{\text{ conjugate axis }}{\text{ transverse axis }}\right)}^2}\end{array}$

Focal Distance of a Point

The difference between the focal distance at any point of the hyperbola is constant and is equal to the length of the transverse axis of the hyperbola.

If $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ is any point on the hyperbola.

$\begin{array}{rl}& \mathrm{SP}=\mathrm{ePM}=e(x_1-\frac{a}{e})=e{x}_1-a\\ & {\mathrm{S}}^{\prime }\mathrm{P}=\mathrm{eP}\mathrm{M}=e(x_1+\frac{a}{e})=e{x}_1+a\\ & |{\mathrm{S}}^{\prime }\mathrm{P}-\mathrm{SP}|=|{\mathrm{ex}}_1+\mathrm{a}-{\mathrm{ex}}_1+\mathrm{a}|=2\mathrm{a}\end{array}$

Length of Latus rectum

Let Latus rectum ${\mathrm{LL}}^{\prime }=2\alpha$
$\mathrm{S}(\mathrm{ae},0)$ is focus, then $\mathrm{LS}={\mathrm{SL}}^{\prime }=\alpha$
Coordinates of L and ${\mathrm{L}}^{\prime }$ become (ae, $\alpha$ ) and (ae, $-\alpha$ ) respectively
Equation of hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\begin{array}{rl}& \therefore \frac{(\mathrm{ae}{)}^2}{{\mathrm{a}}^2}-\frac{{\alpha }^2}{{\mathrm{b}}^2}=1\Rightarrow {\alpha }^2={\mathrm{b}}^2({\mathrm{e}}^2-1)\\ & {\alpha }^2={\mathrm{b}}^2\left(\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\right)[{\mathrm{b}}^2={\mathrm{a}}^2({\mathrm{e}}^2-1)]\\ & \alpha =\frac{{\mathrm{b}}^2}{\mathrm{a}}\\ & \Rightarrow 2\alpha ={\mathrm{LL}}^{\prime }=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\end{array}$

End-points of a latus rectum

For $LR$ passing through $\mathrm{S}(\mathrm{ae},0)$ :

$L(ae,\frac{b^2}{a});L^{\prime }(ae,-\frac{b^2}{a})$
For LR passing through $S^{\prime }(-ae,0)$ :

$L_1(-ae,\frac{b^2}{a});L_1^{\prime }(-ae,-\frac{b^2}{a})$

Parametric equation of Hyperbola

The equations $x=a\sec \theta ,y=b\tan \theta$ are called the parametric equation of the hyperbola
The circle with centre $O(0,0)$ and $OA$ as the radius is called the auxiliary circle of the hyperbola.

Derivation of Parametric equation of Hyperbola

Draw $PN$ perpendicular to the $x$-axis axis and $NQ$ be a tangent to the auxiliary circle. Let be $\mathrm{\angle }\mathrm{QON}=\theta$ (This angle is also known as Eccentric Angle). Hence, the parametric equation of circle at point $Q(a\cos \theta$, a \sin \theta )$.

$\begin{array}{rl}& \text{ now, }x=\frac{\mathrm{ON}}{\mathrm{OQ}}\cdot \mathrm{OQ}=\sec \theta \cdot \mathrm{a}\\ & \mathrm{x}=\mathrm{a}\sec \theta \\ & \mathrm{P}=(\mathrm{a}\sec \theta ,\mathrm{y})\end{array}$
$P$ lies on the hyperbola

$\begin{array}{rl}& \frac{a^2{\sec }^2\theta }{a^2}-\frac{y^2}{b^2}=1\\ & y=\pm b\tan \theta \end{array}$
Point $P$ is $(\mathrm{a}\sec \theta ,\mathrm{b}\tan \theta )$

Conjugate Hyperbola

Conjugate hyperbolas are hyperbolas with the same center but the transverse axes and the conjugate axes of the hyperbolas are interchanged.

For instance, the conjugate hyperbola of the hyperbola $\frac{(x-2{)}^2}{4}-\frac{(y-3{)}^2}{9}=1$ is $\frac{(x-2{)}^2}{9}-\frac{(y-3{)}^2}{4}=1$.

Rectangular Hyperbola

A rectangular hyperbola is a special type of hyperbola whose asymptotes of hyperbola are perpendicular to each other. And the length of the conjugate axis is equal to transverse axis.

In this case, $a=b$.

Rectangular Hyperbola Shape

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

The equation of a rectangular hyperbola can be denoted using various forms, as per the orientation and center entails 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 form of the rectangular hyperbola equation:

$xy=c^2$.

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, is:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Parametric Form

The parametric form of rectangular parabola equation 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 a Rectangular Hyperbola with equation $xy=c^2$ where $c$ is a constant that determines the scale of the hyperbola.

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
$$\begin{array}{rl}& xy=\frac{a^2}{2}\\ & \Rightarrow xy=c^2\end{array}$$

For rectangular hyperbola, $xy=c^2$

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 $={\mathrm{AA}}^{\prime }=2\sqrt{2}c$

Parabola vs Hyperbola

Both parabola and hyperbola are conics. The difference between hyperbola and parabola are

Recommended Video Based on Hyperbola

Solved Examples Based on Hyperbola

Example 1: Let $H$ be the hyperbola, whose foci are $(1\pm \sqrt{2},0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is
[JEE MAINS 2023]
Solution

$\begin{array}{rl}& {\mathrm{F}}_1{\mathrm{F}}_2=2\mathrm{ae}=(1+\sqrt{2})-(1-\sqrt{2})=2\sqrt{2}\\ & \mathrm{ae}=\sqrt{2}\\ & \mathrm{e}=\sqrt{2}\\ & \Rightarrow \mathrm{a}=1\Rightarrow \mathrm{b}=1(\because \mathrm{e}=\sqrt{2})\\ & \text{ L.L.R. }=\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{2(1{)}^2}{1}=2\end{array}$

Hence, the answer is 2

Example 2: Let the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{7}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{\alpha }=\frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is:
[JEE MAINS 2022]
Solution: For the ellipse are

$\begin{array}{rl}& =4\sqrt{1-\frac{7}{16}}\\ & =4\times \frac{3}{4}\\ & =3\end{array}$
For the hyperbola, ae should be 3.

$\begin{array}{rl}& \frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{\alpha }{25}\right)}=1\\ & \Rightarrow \frac{12}{5}\cdot \sqrt{1+\frac{\alpha }{144}}=3\\ & \Rightarrow \sqrt{\frac{144+\alpha }{144}}=\frac{15}{12}\\ & \Rightarrow 144+\alpha =225\\ & \Rightarrow \alpha =81\end{array}$

Latus rectum $=\frac{2b^2}{\mathrm{a}}$

$=\frac{2\cdot 81}{25\cdot \frac{12}{5}}=\frac{27}{10}$
Hence, the answer is $\frac{27}{10}$

Example 3: If the line $\mathrm{x}-1=0$ is a directrix of the hyperbola ${\mathrm{kx}}^2-{\mathrm{y}}^2=6$, then the hyperbola passes through the point
[JEE MAINS 2022]
Solution: ${\mathrm{kx}}^2-{\mathrm{y}}^2=6$

$\begin{array}{rl}& \Rightarrow \frac{x^2}{\left(\frac{6}{k}\right)}-\frac{y^2}{6}=1\\ & a^2=\frac{6}{k},b^2=6\\ & e^2=1+\frac{b^2}{a^2}=1+\frac{6k}{6}=1+k\end{array}$

Directrix $\Rightarrow \mathrm{x}=\frac{\mathrm{a}}{\mathrm{e}}$

$\Rightarrow x=\frac{\sqrt{6}}{\sqrt{\mathrm{k}}\sqrt{1+\mathrm{k}}}$

Given that this equals 1.

$\begin{array}{rl}& \frac{\sqrt{6}}{\sqrt{\mathrm{k}(1+\mathrm{k})}}=1\\ & \Rightarrow {\mathrm{k}}^2+\mathrm{k}-6=0\\ & \Rightarrow \mathrm{k}=2\\ & 2{\mathrm{x}}^2-{\mathrm{y}}^2=6\end{array}$

$(\sqrt{5},-2)$ satisfies it
Hence, the answer is $(\sqrt{5},-2)$

Example 4: Let $\mathrm{H}:\frac{x^2}{{\mathrm{a}}^2}-\frac{y^2}{{\mathrm{b}}^2}=1,\mathrm{a}>0,\mathrm{b}>0$, be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4(2\sqrt{2}+\sqrt{14})$. If the eccentricity $H$ is $\frac{\sqrt{11}}{2}$, then the value of ${\mathrm{a}}^2+{\mathrm{b}}^2$ is equal to [JEE MAINS 2022]

Solution : Given: $2\mathrm{a}+2\mathrm{b}=4(2\sqrt{2}+\sqrt{14})$

$\mathrm{a}+\mathrm{b}=2(2\sqrt{2}+\sqrt{14})$
Now, ${\mathrm{e}}^2=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}$

$\Rightarrow \frac{11}{4}=1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}$
$\begin{array}{rl}& \Rightarrow \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{7}{4}\\ & \Rightarrow \mathrm{b}=\frac{\sqrt{7}}{2}\mathrm{a}\end{array}$
Using this in (i), we get:

$\begin{array}{rl}& \left(\frac{\sqrt{7}+2}{2}\right)\mathrm{a}=2\sqrt{2}(2+\sqrt{7})\\ & \mathrm{a}=4\sqrt{2}\\ & \Rightarrow {\mathrm{a}}^2=32\\ & \Rightarrow {\mathrm{b}}^2=\frac{7}{4}{\mathrm{a}}^2=56\\ & \Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=32+56=88\end{array}$
Hence, the answer is $88$

Example 5 : Let $\mathrm{a}>0,\mathrm{b}>0$. Let $e$ and $l$ respectively be the eccentricity and length of the latus rectum of the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ Let ${\mathrm{e}}^{\prime }$ and $l^{\prime }$ respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If
${\mathrm{e}}^2=\frac{11}{14}l$ and ${\left({\mathrm{e}}^{\prime }\right)}^2=\frac{11}{8}{l}^{\prime }$, then the value of $77\mathrm{a}+44\mathrm{b}$ is equal to :
[JEE MAINS 2022]

Solution:
Given ${\mathrm{e}}^2=\frac{11}{14}\mathrm{l}$

$\begin{array}{rl}& \Rightarrow 1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{11}{14}\cdot \left(\frac{2{\mathrm{b}}^2}{\mathrm{a}}\right)\\ & \Rightarrow \frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{22{\mathrm{b}}^2}{14\mathrm{a}}\\ & \Rightarrow 7({\mathrm{a}}^2+{\mathrm{b}}^2)=11{\mathrm{b}}^2\mathrm{a}\end{array}$
Also ${\mathrm{e}}^{\prime 2}=\frac{11}{8}{\mathrm{l}}^{\prime }$

$\begin{array}{rl}& \Rightarrow 1+\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}=\frac{11}{8}\cdot \frac{2{\mathrm{a}}^2}{\mathrm{b}}\\ & \Rightarrow {\mathrm{b}}^2+{\mathrm{a}}^2={\mathrm{b}}^2\left(\frac{11{\mathrm{a}}^2}{4\mathrm{b}}\right)\\ & \Rightarrow 4({\mathrm{a}}^2+{\mathrm{b}}^2)=11{\mathrm{a}}^2\mathrm{b}\end{array}$
Divide eqn (i) and (ii)

$\frac{7}{4}=\frac{b}{a}$
Also $4({\mathrm{a}}^2+{\mathrm{b}}^2)=11{\mathrm{a}}^2\mathrm{b}$
$\Rightarrow 4(1+{\left(\frac{\mathrm{b}}{\mathrm{a}}\right)}^2)=11\mathrm{b}$
$\begin{array}{rl}& \Rightarrow 11\mathrm{b}=4(1+\frac{49}{16})\\ & \Rightarrow 11\mathrm{b}=4\left(\frac{65}{16}\right)\\ & \Rightarrow \mathrm{b}=\frac{65}{44}\\ & \Rightarrow \mathrm{a}=\frac{4\mathrm{b}}{7}=\frac{4}{7}\cdot \frac{65}{44}=\frac{65}{77}\\ & \therefore 77\mathrm{a}+44\mathrm{b}=65+65=130\end{array}$
Hence, the answer is $130$.

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