Asymptotes of Hyperbolas: Meaning, Formula, Examples

The Asymptote of a curve is a straight line such that the distance between the curve and the line approaches to zero when one or both of the $ x $ - and $ y $ - coordinates approach infinity. An asymptote of any hyperbola is a straight line that touches it at infinity. In real life, we use asymptotes in designing airplanes and designing objects.

This Story also Contains

1. What are Asymptotes of Hyperbola?
2. Equation of Asymptotes of Hyperbola
3. Derivation of Asymptotes of Hyperbola
4. Angle Between Asymptotes of Hyperbola
5. Solved Examples Based on Asymptotes of Hyperbolas

In this article, we will cover the concept of the Asymptotes 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 twenty-one questions have been asked on JEE MAINS( 2013 to 2023) from this topic.

What are Asymptotes of Hyperbola?

The Asymptote of a curve is a straight line such that the distance between the curve and the line approaches to zero when one or both of the x - and y - coordinates approach infinity.
For example, the Asymptote of the curve $y=1 / x$ is a straight line $y=0$ and $x=0$.

An asymptote of any hyperbola is a straight line that touches it at infinity.

Equation of Asymptotes of Hyperbola

The equation of the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $y= \pm \frac{b}{a} x$ or $\frac{x}{a} \pm \frac{y}{b}=0$

Derivation of Asymptotes of Hyperbola

To find the asymptotes of the hyperbola,
Let the straight line $y=m x+c$ is asymptotes to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
put the value of $y$ in the Eq. of hyperbola

$
\begin{aligned}
& \frac{x^2}{a^2}-\frac{(m x+c)^2}{b^2}=1 \\
& \left(a^2 m^2-b^2\right) x^2+2 a^2 m c x+a^2\left(c^2+b^2\right)=0
\end{aligned}
$

since asymptotes touch the hyperbola at infinity, so both roots of the quadratic equation must be infinite and the condition for which is the coefficient of $x^2$ and $x$ must be zero.

$
\begin{array}{ll}
\therefore & a^2 m^2-b^2=0 \\
\text { and } & 2 a^2 m c=0
\end{array}
$
$
\mathrm{m}= \pm \frac{\mathrm{b}}{\mathrm{a}} \text { and } \mathrm{c}=0
$

put the value of $m$ in $y=m x+c$

$
y= \pm \frac{b}{a} x \quad \text { or } \quad \frac{x}{a} \pm \frac{y}{b}=0
$

Angle Between Asymptotes of Hyperbola

The angle between the asymptotes of the hyperbola $\frac{y^2}{a^2}-\frac{y^2}{b^2}=1$ is $2 \tan ^{-1}\left(\frac{b}{a}\right)$
If the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $2 \theta$ then $e=\sec \theta$

Solved Examples Based on Asymptotes of Hyperbolas

Example 1: The angle between the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $60^{\circ}$ and the product of the perpendicular drawn from the foci upon its any tangent is 9 , then the locus of the point of intersection of perpendicular tangents of the hyperbola can be
Solution: The angle between the asymptotes of the hyperbola $\frac{y^2}{a^2}-\frac{y^2}{b^2}=1$ is $2 \tan ^{-1}\left(\frac{b}{a}\right)$
If the angle between the asymptotes of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $2 \theta$ then $e=\sec \theta$
Angle between asymptotes

$
\begin{aligned}
& 2 \tan ^{-1} \frac{b}{a}=60^0 \\
& \tan ^{-1} \frac{b}{a}=30^0 \\
& \frac{b}{a}=\frac{1}{\sqrt{3}} \\
& a^2=3 b^2
\end{aligned}
$
$
\begin{aligned}
& \because b^2=9 \\
& \Rightarrow a^2=27
\end{aligned}
$

$\therefore$ required locus is $x^2+y^2=27-9=18$
Hence, the answer is $x^2+y^2=18$

Example 2: From a point $\mathrm{P}(1,2)$ pair of tangents are drawn to a hyperbola ' H ' where the two tangents touch different arms of hyperbola. Equation of asymptotes of hyperbola H are $\sqrt{3} x-y+5=0$ and $\sqrt{3} x+y-1=0$ then eccentricity of ' H ' is:

Solution: Equation of angle bisectors -

$
\frac{a_1 x+b_1 y+c_1}{\sqrt{a_1^2+b_1^2}}=\frac{ \pm a_2 x+b_2 y+c_2}{\sqrt{a_2^2+b_2^2}}
$

Angle bisectors of the lines $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$
Acute and obtuse angle bisectors -
If $|\tan \Theta|<1$, bisector is acute angle bisector. If $|\tan \Theta|>1$, bisector is obtuse angle bisector.
It $\Theta$ is the angle between one of the lines and one of the bisectors. the origin lies in the acute angle of asymptotes.
$P(1,2)$ lies in obtuse angle of asymptotes
The acute angle between the asymptotes is

Hence, the answer is $2 \sqrt{3}$

Example 3: The angle between the asymptotes of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is equal to:
Solution: Asymptotes of the given hyperbola $y= \pm \frac{b x}{a}$.
Therefore, the angle between them is

$
2 \tan ^{-1}\left(\frac{b}{a}\right)
$
Hence, the answer is

$
2 \tan ^{-1}\left(\frac{b}{a}\right)
$

Example 4: The asymptotes to the hyperbola $x y=h x+h y$ are:
Solution: The equation of asymptotes is

$
x y-h x-k y+\lambda=0
$

where, $\lambda$ is a constant and Eq. (i) represents a pair of straight lines.
Here, $\quad A=0, B=0, C=\lambda, 2 H=1,2 G=-h$ and $2 F=-k$
Then, $\quad A B C+2 F G H-A F^2-B G^2-C H^2=0$
$
\begin{aligned}
& \Rightarrow 0+2\left(-\frac{k}{2}\right)\left(-\frac{h}{2}\right)\left(\frac{1}{2}\right)-0-0-\lambda \cdot \frac{1}{4}=0 \\
& \Rightarrow \quad \frac{h k}{4}=\frac{\lambda}{4} \Rightarrow \lambda=h k
\end{aligned}
$
On putting $\lambda=h k$ in Eq. (i), we get

$
\begin{aligned}
& \Rightarrow \quad x y-h x-k y+h k=0 \\
& \Rightarrow \quad(x-k)(y-h)=0
\end{aligned}
$
So, the asymptotes are $x=k$ and $y=h$
Hence, the answer is $x=k$ and $y=h$

Example 5: If ' $e$ ' is the eccentricity of the hyperbola $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ and ' $\theta$ ' be angle between its asymptotes, then $\cos \frac{\theta}{2}$ is equal to
Solution: If ' $\theta$ ' be the angle between asymptotes, then

$
\begin{aligned}
& \tan \frac{\theta}{2}=\frac{b}{a} \Rightarrow \cos \frac{\theta}{2} \\
& =\frac{a}{\sqrt{a^2+b^2}}=\frac{1}{\sqrt{1+\frac{b^2}{a^2}}}=\frac{1}{e}
\end{aligned}
$
Hence, the answer is $1 / \mathrm{e}$