Tangents to Hyperbolas: Equation, Formula, Examples

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