Intersection of Line and the Hyperbola

A line may meet the Hyperbola in one point or two distinct points or it may not meet the Hyperbola at all. If the line meets the Hyperbola at one point is called Tangent and If the line meets the hyperbola at two points it is called a chord. In real life, we use tangents in the construction and navigation field to calculate distances, heights, and angles.

This Story also Contains

1. What is the Hyperbola?
2. Equation of Hyperbola
3. Intersection of Hyperbola and Line
4. Solved Examples Based on Line and the Hyperbola

In this article, we will cover the concept of Line and 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 twelve questions have been asked on JEE MAINS( 2013 to 2023) from this topic including two in 2022.

What is the Hyperbola?

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 e > 1.

Equation of Hyperbola

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

Intersection of Hyperbola and Line

Hyperbola: $\quad \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Line: $y=m x+c$
After solving Eq. (i) and Eq. (ii)

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

The above equation is quadratic in $x$

The line will cut the hyperbola in two points may be real, coincident or imaginary, depending on the value of Discriminant, D.

Case 1: If D > 0, then two real and distinct roots which means two real and distinct points of intersection of the line and the hyperbola. In this case, the line is secant (chord) to the hyperbola.

Case 2: If D = 0, then equal real roots which means the line is tangent to the hyperbola. Solving D = 0 we get the condition for tangency, which is $c^2=a^2 m^2-b^2$

Case 3: If D < 0, then no real root which means the line and hyperbola do not intersect.

Solved Examples Based on Line and the Hyperbola

Example 1: Let the eccentricity of the hyperbola $H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be $\sqrt{\frac{5}{2}}$ and length of its latus rectum be $6 \sqrt{2}$, If $\mathrm{y}=2 \mathrm{x}+\mathrm{c}$ is a tangent to the hyperbola H, then the value of $\mathrm{c}^2$ is equal to
[JEE MAINS 2022]
Solution

$\begin{aligned}
& \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{5}{2} \Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{3}{2} \quad--(1) \\
& \frac{2 \mathrm{~b}^2}{\mathrm{a}^2}=6 \sqrt{2} \Rightarrow 2 \mathrm{~b}^2=6 \sqrt{2} \mathrm{a} \Rightarrow 3 \mathrm{a}^2=6 \sqrt{2} \mathrm{a} \Rightarrow \mathrm{a}=2 \sqrt{2}, \mathrm{~b}=2 \sqrt{3}
\end{aligned}$

The equation of the tangent of the slope $\mathrm{m}=2$ is

$\begin{aligned}
& y=2 x+\sqrt{a^2 \times 2^2-b^2} \\
& y=2 x+\sqrt{32-12} \quad c=\sqrt{20} \Rightarrow c^2=20
\end{aligned}$
Hence, the answer is 20

Example 2: Let $\lambda x-2 y=\mu$ be a tangent to the hyperbola $\mathrm{a}^2 \mathrm{x}^2-\mathrm{y}^2=\mathrm{b}^2$. Then $\left(\frac{\lambda}{\mathrm{a}}\right)^2-\left(\frac{\mu}{\mathrm{b}}\right)^2$ is equal to:
[JEE MAINS 2022]
Solution

$\frac{x^2}{\left(\frac{b^2}{a^2}\right)}-\frac{y^2}{b^2}=1$
Tangent: $\quad y=\frac{\lambda}{2} x-\frac{\mu}{2}$
Using condition for tangency for hyperbola:

$\begin{aligned}
& \frac{\mu^2}{4}=\frac{b^2}{\mathrm{a}^2} \times \frac{\lambda^2}{4}-b^2 \\
& \frac{\mu^2}{4 \mathrm{~b}^2}=\frac{1}{4}\left(\frac{\lambda^2}{\mathrm{a}^2}-4\right) \\
& \frac{\lambda^2}{\mathrm{a}^2}-\frac{\mu^2}{\mathrm{~b}^2}=4
\end{aligned}$
Hence, the answer is 4

Example 3: The locus of a point $P(\alpha, \beta)$ moving under the condition that the line $y=\alpha x+\beta$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
Solution: Condition for Tangency in Hyperbola - $C^2=a^2 m^2-b^2$
For the Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$y=\alpha x+\beta$ is tangent to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Condition for tangency for $\mathrm{y}=\mathrm{mx}+\mathrm{C}$ is $\mathrm{C}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2$
i.e. $\beta^2=a^2 \alpha^2-b^2$

Thus replace $(\alpha, \beta)$ by $(x, y)$
$y^2=a^2 x^2-b^2$ which is hyperbola.
Hence, the answer is a hyperbola

Hence, the answer is a hyperbola

Example 4: If $P\left(\theta_1\right)$ and $Q\left(\theta_2\right)$ are the extremities of any focal chord of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $\cos ^2 \frac{\theta_1+\theta_2}{2}=\lambda \cos ^2 \frac{\theta_1-\theta_2}{2}$, where $\lambda$ is equal to
Solution: Equation of any chord joining the points $P\left(\theta_1\right)$ and $Q\left(\theta_2\right)$ is, $\frac{\mathrm{x}}{\mathrm{a}} \cos \left(\frac{\theta_1-\theta_2}{2}\right)-\frac{\mathrm{y}}{\mathrm{b}}$
$\sin \left(\frac{\theta_1+\theta_2}{2}\right)=\cos \left(\frac{\theta_1+\theta_2}{2}\right)_{\text {If it passes through (ae, } 0 \text { ), then }}$

$\begin{aligned}
& \Rightarrow \mathrm{e}^2 \cos ^2\left(\frac{\theta_1-\theta_2}{2}\right)=\cos ^2\left(\frac{\theta_1+\theta_2}{2}\right) \\
& \Rightarrow \lambda=\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2}
\end{aligned}$
Hence, the answer is $\frac{a^2+b^2}{a^2}$

Example 5: If $(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta)$ and $(\operatorname{asec} \phi, \mathrm{b} \tan \phi)$ be the coordinate of the ends of a focal chord of $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$, then $\tan \frac{\theta}{2} \tan \frac{\phi}{2}$ equals to
Solution: The equation of the chord connecting the points $(\operatorname{asec} \theta, \mathrm{b} \tan \theta)$ and $(\operatorname{asec} \phi, \mathrm{b} \tan \phi)$ is

$\frac{\mathrm{x}}{\mathrm{a}} \cos \left(\frac{\theta+\phi}{2}\right)-\frac{\mathrm{y}}{\mathrm{b}} \sin \left(\frac{\theta+\phi}{2}\right)=\cos \left(\frac{\theta-\phi}{2}\right)$

If it passes through $(\mathrm{ae}, 0)$; we have, $\cos \left(\frac{\theta-\phi}{2}\right)=\cos \left(\frac{\theta+\phi}{2}\right)$

$\mathrm{e}=\frac{\cos \left(\frac{\theta+\phi}{2}\right)}{\cos \left(\frac{\theta-\phi}{2}\right)}=\frac{1-\tan \frac{\theta}{2} \cdot \tan \frac{\phi}{2}}{1+\tan \frac{\theta}{2} \tan \frac{\phi}{2}} \Rightarrow \tan \frac{\theta}{2} \cdot \tan \frac{\phi}{2}=\frac{1-\mathrm{e}}{1+\mathrm{e}}$
Hence, the answer is $\frac{1-\mathrm{e}}{1+\mathrm{e}}$