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

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

Derivation of Equation of Hyperbola

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

$
\begin{aligned}
& \frac{P S}{P M}=e \Rightarrow(P S)^2=e^2(P M)^2 \\
& (x-a e)^2+(y-0)^2=e^2\left(x-\frac{a}{e}\right)^2 \\
& x^2+a^2 e^2-2 a e x+y^2=e^2 x^2-2 a e x+a^2 \\
& x^2\left(e^2-1\right)-y^2=a^2\left(e^2-1\right) \\
& \frac{x^2}{a^2}-\frac{y^2}{a^2\left(e^2-1\right)}=1 \\
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \quad b^2=a^2\left(e^2-1\right)
\end{aligned}
$

Important Terms related to Hyperbola

• Centre: All chord passing through point $O$ is bisected at point $O$. Here $O$ is the origin, i.e. $(0,0)$.
• 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'M' are two directrices of the hyperbola and their equations are $x=$ ae and $x=-a e$.
• Double Ordinate: If a line perpendicular to the transverse axis of the hyperbola meets the curve at Q and Q', then QQ' 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.

Formula of Eccentricity

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

$
\begin{aligned}
& \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right) \\
& \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{aligned}
$

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

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

Length of Latus rectum

Let Latusrectum $\mathrm{LL}^{\prime}=2 \alpha$
$\mathrm{S}(\mathrm{ae}, 0)$ 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, $\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

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

End-points of a latus rectum|
For LR passing through $\mathrm{S}(\mathrm{ae}, 0)$ :

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

$
L_1\left(-a e, \frac{b^2}{a}\right) ; L_1^{\prime}\left(-a e,-\frac{b^2}{a}\right)
$
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 $O A$ 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 $N Q$ be a tangent to the auxiliary circle. Let be $\angle \mathrm{QON}=\theta$ (This angle is also known as Eccentric Angle). Hence, the parametric equation of circle at point Q (a cos $\theta$, a sin $\theta$ ).

$
\begin{aligned}
& \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{aligned}
$
P lies on the hyperbola

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

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{aligned}
& \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{aligned}
$

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

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

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

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

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{aligned}
& \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{aligned}
$

$(\sqrt{5},-2) \quad$ 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{aligned}
& \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{7}{4} \\
& \Rightarrow \mathrm{b}=\frac{\sqrt{7}}{2} \mathrm{a}
\end{aligned}
$
Using this in (i), we get:

$
\begin{aligned}
& \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{aligned}
$
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{aligned}
& \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\left(\mathrm{a}^2+\mathrm{b}^2\right)=11 \mathrm{~b}^2 \mathrm{a}
\end{aligned}
$
Also $\mathrm{e}^{\prime 2}=\frac{11}{8} \mathrm{l}^{\prime}$

$
\begin{aligned}
& \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\left(\mathrm{a}^2+\mathrm{b}^2\right)=11 \mathrm{a}^2 \mathrm{~b}
\end{aligned}
$
Divide eqn (i) and (ii)

$
\frac{7}{4}=\frac{b}{a}
$
Also $4\left(\mathrm{a}^2+\mathrm{b}^2\right)=11 \mathrm{a}^2 \mathrm{~b}$
$
\Rightarrow 4\left(1+\left(\frac{\mathrm{b}}{\mathrm{a}}\right)^2\right)=11 \mathrm{~b}
$
$
\begin{aligned}
& \Rightarrow 11 \mathrm{~b}=4\left(1+\frac{49}{16}\right) \\
& \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{aligned}
$
Hence, the answer is 130.