Equation of Normal to Hyperbola

What is 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).
OR,
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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \quad$ where $b^2=a^2\left(e^2-1\right)$

Equation of Normal of Hyperbola in Point form

The equation of normal at $\left(x_1, y_1\right)$ to the hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2$

Derivation of Equation of Normal of Hyperbola in Point form

We know that the equation of tangent in point form at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$

$
\frac{x x_1}{a^2}-\frac{y y_1}{b^2}=1
$
Slope of tangent at $\left(x_1, y_1\right)$ is $\frac{b^2 x_1}{a^2 y_1}$
$\therefore \quad$ Slope of normal at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is $-\frac{\mathrm{a}^2 \mathrm{y}_1}{\mathrm{~b}^2 \mathrm{x}_1}$
Hence, the equation of normal at point $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ is

$
\left(\mathrm{y}-\mathrm{y}_1\right)=-\frac{\mathrm{a}^2 \mathrm{y}_1}{\mathrm{~b}^2 \mathrm{x}_1}\left(\mathrm{x}-\mathrm{x}_1\right)
$
$
\text { or } \quad \frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2
$
Equation of Normal of Hyperbola in Parametric form

The equation of normal at $(a \sec \theta, b \tan \theta)$ to the hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $a x \cos \theta+b y \cot \theta=a^2+b^2$

Derivation of Equation of Normal of Hyperbola in Parametric form

The equation of normal in point form is $\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2$

$
\begin{aligned}
& \text { Put }\left(x_1, y_1\right) \equiv(a \sec \theta, b \tan \theta) \\
& \Rightarrow \quad \frac{a^2 x}{a \sec \theta}+\frac{b^2 y}{b \tan \theta}=a^2+b^2 \\
& \Rightarrow \quad a x \cos \theta+b y \cot \theta=a^2+b^2
\end{aligned}
$
The equation of normal at $(a \sec \theta, b \tan \theta)$ to the hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $a x \cos \theta+b y \cot \theta=a^2+b^2$

The equation of normal in point form is $\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2$ put $\left(x_1, y_1\right) \equiv(a \sec \theta, b \tan \theta)$.

$
\begin{aligned}
& \Rightarrow \quad \frac{a^2 x}{a \sec \theta}+\frac{b^2 y}{b \tan \theta}=a^2+b^2 \\
& \Rightarrow \quad a x \cos \theta+b y \cot \theta=a^2+b^2
\end{aligned}
$

Equation of Normal of Hyperbola in Slope form

The equation of normal of slope m to the hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are $y=m x \mp \frac{m\left(a^2+b^2\right)}{\sqrt{a^2-m^2 b^2}}$ and coordinate of point of contact is $\left( \pm \frac{a^2}{\sqrt{a^2-m^2 b^2}}, \mp \frac{m b^2}{\sqrt{a^2-m^2 b^2}}\right)$

The equation of normal at $(a \sec \theta, b \tan \theta)$ to the hyperbola, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $a x \cos \theta+b y \cot \theta=a^2+b^2$

Derivation of Equation of Normal of Hyperbola in Slope form

$
\Rightarrow \quad y=-\frac{a \sin \theta}{b} x+\frac{a^2+b^2}{b \cot \theta}
$
Let, $\quad-\frac{\mathrm{a} \sin \theta}{\mathrm{b}}=\mathrm{m}$

Hence, the equation of normal becomes

$
y=m x \mp \frac{m\left(a^2+b^2\right)}{\sqrt{a^2-m^2 b^2}}, \text { where } m \in\left[-\frac{a}{b}, \frac{a}{b}\right]
$
Pair of Tangents

$
\begin{aligned}
& \therefore \quad \sin \theta=-\frac{b m}{a} \\
& \therefore \quad \cot \theta= \pm \frac{\sqrt{a^2-b^2 m^2}}{b m}
\end{aligned}
$

The combined equation of pair of tangents from the point $P\left(x_1, y_1\right)$ to the hyperbola

$
\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is }\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1\right)\left(\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}-1\right)=\left(\frac{x x_1}{a^2}-\frac{y y_1}{b^2}-1\right)^2 \\
& \text { or, } \quad S S_1=T^2 \\
& \text { where, } \quad \mathrm{S}=\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}-1 \\
& \mathrm{~S}_1=\frac{\mathrm{x}_1^2}{\mathrm{a}^2}-\frac{\mathrm{y}_1^2}{\mathrm{~b}_2^2}-1 \\
& \mathrm{~T}=\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}-1
\end{aligned}
$
Note:
The formula $\mathrm{SS}_1=\mathrm{T}^2$ can be used to find the combined equation of a pair of tangents for any general hyperbola as well.
Chord of Contact

The equation of chord of contact of tangents from the point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{x_1}{a^2}-\frac{y_1}{b^2}=1$
i.e. $T=0$ which is a chord of contact $Q R$.

Equation of Chord bisected at a given point

The equation of chord of the hyperbola $\frac{\kappa}{\mathrm{a}^2}-\frac{\sqrt{\mathrm{b}^2}}{}=1$ bisected at a given point $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$
is $\frac{\mathrm{xx}_1}{\mathrm{a}^2}-\frac{\mathrm{yy}_1}{\mathrm{~b}^2}-1=\frac{x_1^2}{\mathrm{a}^2}-\frac{y^2}{\mathrm{~b}^2}-1$
or, $\mathrm{T}=\mathrm{S}_1$

Solved Examples Based on Equation of Normal of Hyperbola

Example 1: Let $\mathrm{m}_1$, and $\mathrm{m}_2$ be the slopes of the tangents drawn from the point $\mathrm{P}(4,1)$ to the hyperbola $\mathrm{H}: \frac{y^2}{25}-\frac{x^2}{16}=1$. If Q is the point from which the tangents are drawn to H have slopes $\left|\mathrm{m}_1\right|$ and $\left|\mathrm{m}_2\right|$ and they make positive intercepts $\alpha$ and $\beta$ on the $x$-axis, then $\frac{(\mathrm{PQ})^2}{\alpha \beta}$ is equal to [JEE MAINS 2023]

Solution: Equation of tangent to the hyperbola $\frac{\mathrm{y}^2}{\mathrm{a}^2}-\frac{\mathrm{x}^2}{\mathrm{~b}^2}=1$
$y=m x \pm \sqrt{a^2-b^2 \mathrm{~m}^2}$
passing through $(4,1)$
$1=4 \mathrm{~m} \pm \sqrt{25-16 \mathrm{~m}^2} \Rightarrow 4 \mathrm{~m}^2-\mathrm{m}-3=0$

Equation of tangent with positive slopes $1 \& \frac{3}{4}$
$\left.\begin{array}{c}4 y=3 x-16 \\ y=x-3\end{array}\right\}$ with positive intercept on $x$-axis.

$
\alpha=\frac{16}{3}, \beta=3
$
Intersection points:

$
\begin{aligned}
& \mathrm{Q}:(-4,-7) \\
& \mathrm{P}:(4,1) \\
& \mathrm{PQ}^2:(128) \\
& \frac{\mathrm{PQ}^2}{\alpha \beta}=\frac{128}{16}=8
\end{aligned}
$
Hence, the answer is 8.

Example 2: Consider a hyperbola $H: x^2-2 y^2=4$. Let the tangent at a point $\mathrm{P}(4, \sqrt{6})$ meet the axis at Q and latus rectum at $\mathrm{R}\left(x_1, y_1\right), x_1>0$. If F is a focus of H which is nearer to the point P, then the area of $\Delta Q F R$ is equal to: [JEE MAINS 2021]

Solution

$
\begin{aligned}
& \frac{x^2}{4}-\frac{y^2}{2}=1 \\
& e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{\frac{3}{2}}
\end{aligned}
$

$\therefore$ Focus $F(a e, 0) \Rightarrow F(\sqrt{6}, 0)$

equation of the tangent at $P$ to the hyperbola is

$
2 x-y \sqrt{6}=2
$

tangent meet $\underline{x}$-axis at $Q(1,0)$
latus rectum $x=\sqrt{6}$ at $R\left(\sqrt{6}, \frac{2}{\sqrt{6}}(\sqrt{6}-1)\right)$
Area of $\Delta \mathrm{QFR}=\frac{1}{2}(\sqrt{6}-1) \cdot \frac{2}{\sqrt{6}}(\sqrt{6}-1)=\frac{(\sqrt{6}-1)^2}{\sqrt{6}}=\frac{7}{\sqrt{6}}-2$
Hence, the answer is $\frac{7}{\sqrt{6}}-2$

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 2020]
Solution

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