Tangents to Hyperbolas: Equation, Formula, Examples

The tangent touches the curve at one point but does not cross it. So the tangent has only one point of contact. The point where the tangent line and the curve meet or intersect is called the point of tangency. In real life, we use tangents in the construction and navigation field to calculate distances, heights, and angles.

This Story also Contains

1. Equation Of Hyperbola
2. Equation of Tangent of Hyperbola in Point Form
3. Equation of Tangent of Hyperbola in Parametric
4. Solved Examples Based on the Equation of Tangent of Hyperbola

In this article, we will cover the concept of the Equation of Tangent to 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 including one in 2017, two in 2019, two in 2020, four in 2021, five in 2022, and four in 2023.

Equation Of 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).

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

Equation of Tangent of Hyperbola in Point Form

The equation of tangent to the hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at point $\left(x_1, y_1\right)$ is $\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=1$

Derivation of Equation of Tangent of Hyperbola in Point Form

Differentiating $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ w.r.t. $x$, we have

$\begin{array}{ll}
& \frac{2 x}{a^2}-\frac{2 y}{b^2} \frac{d y}{d x}=0 \\
\Rightarrow \quad & \frac{d y}{d x}=\frac{b^2 x}{a^2 y} \\
\Rightarrow \quad & \left(\frac{d y}{d x}\right)_{(x, y)}=\frac{b^2 x_1}{a^2 y_1}
\end{array}$
Hence, equation of the tangent is $y-y_1=\frac{b^2 x_1}{a^2 y_1}\left(x-x_1\right)$ or

$\frac{x x_1}{a^2}-\frac{y y_1}{b^2}=\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}$
But $\left(x_1, y_1\right)$ lies on the hyperbola $\Rightarrow \frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}=1$
Hence, the equation of the tangent is

$\begin{aligned}
& \quad \frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1 \\
& \text { or } \quad \frac{x x_1}{a^2}-\frac{y y_1}{b^2}-1=0 \text { or } T=0 \\
& \text { where } \quad T=\frac{x x_1}{a^2}-\frac{y y_1}{b^2}-1
\end{aligned}$

Note:

T = 0 can be used to get the equation of tangent on the point (x₁, y₁) lying on any general hyperbola as well.

Equation of Tangent of Hyperbola in Parametric

The equation of tangent to the hyperbola, $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ at $(\mathrm{a} \sec \theta, \mathrm{b} \tan \theta)$ is $\frac{\mathrm{x}}{\mathrm{a}} \sec \theta-\frac{\mathrm{y}}{\mathrm{b}} \tan \theta=1$
(This can easily be derived by putting $\mathrm{x}_1=\mathrm{a} \sec \theta$ and $\mathrm{y}_1=\mathrm{b} \tan \theta$ in the point form of tangent)

Equation of Tangent of Hyperbola in Slope Form

We have studied that if the line $y=m x+c$ is tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $c^2=a^2 m^2-b^2$. So the equation of tangent is $y=m x \pm \sqrt{a^2 m^2-b^2}$.

These equations are equations of two parallel tangents to hyperbola having slope m.

Solved Examples Based on the Equation of Tangent of Hyperbola

Example 1 : The foci of a hyperbola are $( \pm 2,0)$ and its eccentricity is $\frac{3}{2}$. A tangent, perpendicular to the line $2 x+3 y=6$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the x and y axes are a and b respectively, then $|6 \mathrm{a}|+|5 \mathrm{~b}|$ is equal to $\qquad$
[JEE MAINS 2023]

Solution

$\begin{aligned}
& 2 \mathrm{ae}=4 \\
& 2 \mathrm{a}\left(\frac{3}{2}\right)=4
\end{aligned}
$

equation of tangent is

$\begin{aligned}
& y=m x \pm \sqrt{a^2 m^2-b^2} \\
& y=\frac{3}{2} x \pm \sqrt{\frac{16}{9}\left(\frac{9}{4}\right)-\frac{20}{9}} \\
& \Rightarrow y=\frac{3 x}{2} \pm \frac{4}{3} \\
& y=0 \Rightarrow a= \pm \frac{8}{9} \\
& x=0 \Rightarrow b= \pm \frac{4}{3} \\
& |6 a|+|5 b|=\frac{16}{3}+\frac{20}{3}=12
\end{aligned}$

Hence, the answer is 12 .

Example 2: Let $\mathrm{P}\left(\mathrm{x}_0, \mathrm{y}_0\right)$ be the point on the hyperbola $3 x^2-4 y^2=36$, which is nearest to the line $3 x+2 y=1$. Then $\sqrt{2}\left(y_0-x_0\right)$ is equal to:
[JEE MAINS 2023]
Solution
We have, $3 x^2-4 y^2=36$ and $3 x+2 y=1$

$\begin{aligned}
& m=-\frac{3}{2} \\
& m=+\frac{3 \sec \theta}{\sqrt{12} \cdot \tan \theta} \\
& \Rightarrow \frac{3}{\sqrt{12}} \times \frac{1}{\sin \theta}=\frac{-3}{2} \\
& \sin \theta=-\frac{1}{\sqrt{3}} \\
& (\sqrt{12} \cdot \sec \theta, 3 \tan \theta) \\
& \left(\sqrt{12} \cdot \frac{\sqrt{3}}{\sqrt{2}},-3 \times \frac{1}{\sqrt{2}}\right) \\
& \Rightarrow\left(\frac{6}{\sqrt{2}}, \frac{-3}{\sqrt{2}}\right)=\left(x_0, y_0\right) \\
& \Rightarrow \sqrt{2}\left(y_0-x_0\right)=\sqrt{2}\left(\frac{-3}{\sqrt{2}}-\frac{6}{\sqrt{2}}\right)=-9
\end{aligned}$

Hence, the answer is -9

Example 3 : The vertices of a hyperbola $H$ are $( \pm 6,0)$ and its eccentricity is $\frac{\sqrt{5}}{2}$. Let N be the normal to H at a point in the first quadrant and parallel to the line $\sqrt{2} x+y=2 \sqrt{2}$. If $d$ is the length of the line segment of $N$ between $H$ and the $y$-axis then $d^2$ is equal to [JEE MAINS 2023]

Solution

Solution

$H: \frac{x^2}{36}-\frac{y^2}{9}=1$

Equation of normal is $6 x \cos \theta+3 y \cot \theta=45$

$\begin{aligned}
& M=-2 \sin \theta=-\sqrt{2} \\
& \theta=\pi / 4
\end{aligned}$

Equation of normal is $\sqrt{2} x+y=15$

$\begin{aligned}
& \mathrm{P}(\operatorname{asec} \theta, b \tan \theta) \\
& \mathrm{P}(6 \sqrt{2}, 3), \mathrm{k}(0,15) \\
& \mathrm{d}^2=216
\end{aligned}$
Hence, the answer is 216

Example 4: Let the focal chord of the parabola $\mathrm{P}: \mathrm{y}^2=4 \mathrm{x}$ along the line $\mathrm{L}: \mathrm{y}=\mathrm{mx}+\mathrm{c}, \mathrm{m}>0$ meet the parabola at the points M and N . Let the line L be a tangent to the hyperbola $\mathrm{H}: \mathrm{x}^2-\mathrm{y}^2=4$. If O is the vertex of P and F is the focus of H on the positive x -axis, then the area of the quadrilateral OMFN is :
[JEE MAINS 2022]

Solution

Line $L$ is tangent to Hyperbola $\frac{x^2}{4}-\frac{y^2}{4}=1$

$\begin{aligned}
& \frac{x^2}{4}-\frac{y^2}{4}=1 \\
& \text { forcus }(a 0,0) \\
& f(2 \sqrt{2}, 0)
\end{aligned}$

line $L: y=m x+$ cpass $(1,0)$

$0=\mathrm{m}+\mathrm{c} \cdots$

$\begin{gathered}
c= \pm \sqrt{a^2 m^2-l^2} \\
c= \pm \sqrt{4 m^2-4} \\
\text { from }(1) \\
-m= \pm \sqrt{4 m^2-4} \\
m^2=4 m^2-4 \\
4=3 m^2 \\
\frac{2}{\sqrt{3}}=m \quad(\text { as } m>0) \\
c=-m \\
c=-\frac{2}{\sqrt{3}} \\
\end{gathered}$

$\begin{aligned}
& y=\frac{2 x}{\sqrt{3}}-\frac{2}{\sqrt{3}} \\
& \mathrm{y}^2=4 \mathrm{x} \\
& \Rightarrow\left(\frac{2 \mathrm{x}-2}{\sqrt{3}}\right)^2=4 \mathrm{x} \\
& \Rightarrow \mathrm{x}^2+1-2 \mathrm{x}=3 \mathrm{x} \\
& x^2-5 x+1=0 \\
& \mathrm{y}^2=4\left(\frac{\sqrt{3} \mathrm{y}+2}{2}\right) \\
& \mathrm{y}^2=2 \sqrt{3} \mathrm{y}+4 \\
& \Rightarrow \mathrm{y}^2-2 \sqrt{3} \mathrm{y}-4=0 \\
& \text { Area }=\left|\frac{1}{2}\right| \begin{array}{ccccc}
0 & x_1 & 2 \sqrt{2} & x_2 & 0 \\
0 & y_1 & 0 & y_2 & 0
\end{array} \| \\
& =\left|\frac{1}{2}\left[-2 \sqrt{2} \mathrm{y}_1+2 \sqrt{2} \mathrm{y}_2\right]\right|=\sqrt{2}\left|\mathrm{y}_2-\mathrm{y}_1\right|=\frac{(\sqrt{2}) \sqrt{12+16}}{111} \\
& =\sqrt{56}=2 \sqrt{14}
\end{aligned}$

Example 5: Let a line $L_1$ be tangent to the hyperbola $\frac{x^2}{16}-\frac{y^2}{4}=1$ and let $L_2$ be the line passing through the origin and perpendicular to $L_1$. If the locus of the point of intersection of $L_1$ and $L_2$ is $\left(x^2+y^2\right)^2=\alpha x^2+\beta y^2$, then $\alpha+\beta$ is equal to
[JEE MAINS 2022]
Solution

$\begin{aligned}
& \frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{4}=1 \\
& \mathrm{~L}_1: \quad \frac{\mathrm{x} \sec \theta}{4}-\frac{\mathrm{y} \tan \theta}{2}=1 \\
& \mathrm{~m}_1=\frac{1}{2 \sin \theta} \\
& \mathrm{L}_2: \quad \mathrm{y}=-2 \sin \theta \mathrm{x}
\end{aligned}$

passes through ( $\mathrm{h}, \mathrm{k}$ )

$\begin{aligned}
& \mathrm{k}=-2(\sin \theta) \mathrm{h}^{\mathrm{h}} \\
& \sin \theta=\frac{-\mathrm{k}}{2 \mathrm{~h}}
\end{aligned}$

from $L_1$ :

$\begin{aligned}
& \frac{\mathrm{h}}{4} \frac{2 \mathrm{~h}}{\sqrt{4 \mathrm{k}^2+\mathrm{k}^2}}-\frac{\mathrm{k}}{2}\left(\frac{-\mathrm{k}}{\sqrt{4 \mathrm{k}^2-\mathrm{k}^2}}\right)=1 \\
& \left(\mathrm{x}^2+\mathrm{y}^2\right)^2=16 \mathrm{x}^2-4 \mathrm{y}^2 \\
& \therefore \quad \alpha=16, \quad \beta=-4 \\
& \therefore \alpha+\beta=12
\end{aligned}$

Hence, the answer is 12