Position of a point with respect to Hyperbola

In this article, we will learn how to determine the position of a point with respect to a hyperbola. The hyperbola drawn on the plane divides the plane into two regions, one is a concave (interior) region where the focus lies and the other is a convex (exterior) region. Specifically, we will cover the three possible locations where a point P(x₁,y₁) can lie - inside, outside or on the hyperbola.

In this article, we will cover the concept of the Position of a Point with Respect to a Hyperbola. This category falls under the broader category of Conic Sections, 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 ten questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2016, and one in 2019.

Position of a point with respect to Hyperbola

In a Hyperbola a point can lie at three positions- inside the hyperbola, outside the hyperbola, and on the hyperbola.

Let P(x₁,y₁) be any point in the plane

(a) P lies outside of the hyperbola : $\frac{\mathrm{x}_1{ }^2}{\mathrm{a}^2}-\frac{\mathrm{y}_1{ }^2}{\mathrm{~b}^2}-1<0$
(b) P lies on of the hyperbola : $\frac{x_1{ }^2}{\mathrm{a}^2}-\frac{y_1{ }^2}{\mathrm{~b}^2}-1=0$
(c) P lies inside of the hyperbola : $\frac{\mathrm{x}_1{ }^2}{\mathrm{a}^2}-\frac{\mathrm{y}_1{ }^2}{\mathrm{~b}^2}-1>0$

Steps to Check the Position of a Point with Respect to Hyperbola

Step 1: Write the equation of the hyperbola in the form

$S = \frac{\mathrm{x}_1{ }^2}{\mathrm{a}^2}-\frac{\mathrm{y}_1{ }^2}{\mathrm{~b}^2}-1$ where $b^2=a^2\left(e^2-1\right)$

Step 2: Substitute the point (x₁, y₁) in S

$S_1 = \frac{x_1^2}{a_1^2}-\frac{y_1^2}{b_1^2}-1$

Step 3:

Case 1: If S₁ > 0, then the point P(x₁, y₁) lies inside the hyperbola.

Case 2: If S₁ = 0, then the point P(x₁,y₁) is on the hyperbola. The value of S₁ at a point on the hyperbola is equal to zero.

Case 3: If S₁ < 0, then the point P(x₁, y₁) lies outside the hyperbola.

Solved Examples Based on the Position of a point with respect to Hyperbola

Example 1: If a directrix of a hyperbola centred at the origin and passing through the point $(4,-2 \sqrt{3})$ is $5 x=4 \sqrt{5}$ and its eccentricity is e, then
[JEE MAINS 2019]
Solution: Equation of directrices- $x= \pm \frac{a}{e}$

For the Hyperbola

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

it passes through $(4,-2 \sqrt{3})$
So,

$\begin{aligned}
& \frac{16}{a^2}-\frac{12}{b^2}=1 \\
& \Rightarrow 16-12 \frac{a^2}{b^2}=a^2
\end{aligned}$

given equation of the directrix

$\begin{aligned}
& 5 x=4 \sqrt{5} \\
& \Rightarrow x=\frac{4}{\sqrt{5}}
\end{aligned}$
Also,

$\begin{aligned}
& x=\frac{a}{e} \\
& \Rightarrow \frac{4}{\sqrt{5}}=\frac{a}{e}
\end{aligned}$

And we know that $b^2=a^2\left(e^2-1\right)$

$\begin{aligned}
& \frac{b^2}{a^2}=e^2-1 \\
& \text { from (1).(2) and (3) } \\
& \Rightarrow 16 e^2-16-12=16 \frac{e^2}{5}\left(e^2-1\right) \\
& \Rightarrow 4 e^4-24 e^2+35=0
\end{aligned}$
Hence, the answer is $4 e^4-24 e^2+35=0$

Example 2: Let a and b respectively be the semi-transverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation 9e²−18e+5=0. If S(5, 0) is a focus and 5x=9 is the corresponding directrix of this hyperbola, then a²−b² is equal to : [JEE MAINS 2016]

Solution: Equation of directrices - $x= \pm \frac{a}{e}$
For the Hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\begin{aligned}
& 9 e^2-18 e+5=0 \\
& \Rightarrow 9 e^2-15 e-3 e+5=0 \\
& \Rightarrow 3 e(3 e-5)-1(3 e-5)=0 \\
& e=\frac{1}{3} \text { or } \\
& e=\frac{5}{3}
\end{aligned}$

for Hyperbola $e=\frac{5}{3}$
Also ae $-\frac{a}{e}=5-\frac{9}{5}=\frac{16}{5}$
$\begin{aligned}
& \frac{5}{3} a-\frac{3}{5} a=\frac{16}{5} \\
& \frac{16 a}{15}=\frac{16}{5} \\
& \Rightarrow a=3
\end{aligned}$
Also,

$\begin{aligned}
& \quad b^2=a^2\left(c^2-1\right)=9\left(\frac{25}{9}-1\right)=16 \\
& \text { Also, } \\
& a^2-b^2=-7
\end{aligned}$
Hence, the answer is -7
Example 3: The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, then the value of $b^2$ is
Solution: Eccentricities of Hyperbola -

$\frac{1}{e_1^2}+\frac{1}{e_2^3}=1$

$e_1$ and $e_2$ are eccentricities of the hyperbola and its conjugate.
For the given ellipse, $a^2=16, b^2=b^2$.

$\therefore e=1-\frac{b^2}{a^2} \Rightarrow e=\frac{\sqrt{16-b^2}}{4}$
Thus the foci of the ellipse are $( \pm a e, 0)$ i.e. $\left( \pm \sqrt{\left(16-b^2\right)}, 0\right)$

Re-arranging the equation of the hyperbola, we get
$\frac{x^2}{(12 / 5)^2}-\frac{y^2}{(9 / 5)^2}=1$
Comparing this with the standard equation, we have
$a=\frac{12}{5}, b=\frac{9}{5}$ and the centre $(0,0)$

$\therefore e^2=1+\left(\frac{b^2}{a^2}\right) \Rightarrow e=\frac{5}{4}$
Thus the foci of the hyperbola are ( $\pm a e, 0)$ i.e. $( \pm 3,0)$
Since the foci of ellipse and hyperbola coincide, therefore

$\sqrt{\left(16-b^2\right)}=3 \Rightarrow b^2=7$
Hence, the answer is 7
Example 4: A hyperbola has its centre at the origin, passes through the point ( 4,2 ), and has a transverse axis of length 4 along the $x$-axis. Then the eccentricity of the hyperbola is
Solution: Let the equation of the required hyperbola be

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Transverse axis length: $2 a=4 \Rightarrow a=2$
So,

$\frac{x^2}{4}-\frac{y^2}{b^2}=1$

Now it passes through $(4,2)$

$\begin{aligned}
& 4-\frac{4}{b^2}=1 \\
& b^2=\frac{4}{3}
\end{aligned}$
Hence, the eccentricity of a hyperbola:

$\begin{aligned}
& \text { Hence, the eccentricity of a hyperbola: } e=\sqrt{1+\frac{b^2}{a^2}} \\
& c=\frac{2}{\sqrt{3}}
\end{aligned}$
Hence, the answer is $c=\frac{2}{\sqrt{3}}$
Example 5: The foci of a hyperbola coincide with the foci of ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$. If the eccentricity of the hyperbola is 3 , then its equation is

Solution

For the ellipse $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ the foci are ( $\pm$ ae, 0 )
. In the present case, $\mathrm{a}=5$.

$\begin{aligned}
& \therefore \quad 9=25\left(1-e^2\right) \Rightarrow e^2=1-\frac{9}{25}=\frac{16}{25} \\
& \therefore \quad e=\frac{4}{5}
\end{aligned}$
Hence, the foci of the ellipse are $( \pm 4,0)$
Let the hyperbola be $\frac{x^2}{a_1^2}-\frac{y^2}{b_1^2}=1$ $\qquad$
And eccentricity $\mathrm{e}_1=3$ (given)

$\therefore \quad b_1^2=a_1^2\left(e_1^2-1\right)=a_1^2(9-1)$

Or $b_1^2=8 a_1^2$
As the foci of the ellipse and the hyperbola coincide, the condition is

$a_1 e_1=a c=4$

Or $a_1 \cdot 3=4$ or $a_1=\frac{4}{3}$
$\therefore$ From (2), $\mathrm{b}_1^2=8 \cdot \frac{16}{9}=\frac{128}{9}$
The required equation of the hyperbola is [by using (1)]

$\frac{x^2}{16}-\frac{y^2}{128}=\frac{1}{9}$

Hence, the answer is 1/9

Summary

Based on the point of contact of the line on the Hyperbola, we can differentiate it. So, knowledge of the point of contact of the line with respect to the Hyperbola is necessary for understanding its properties and characteristics. Understanding of point of contact of the line on the Hyperbola is necessary for solving theoretical as well as real-life-based problems.