Equation of Normal to 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 line perpendicular to the tangent to the curve at the point of contact is normal to the Hyperbola. In real life, we use Hyperbolas in race tracks, architectural design, mirrors, and celestial orbits.

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This Story also Contains

1. What is Hyperbola?
2. Equation of Normal of Hyperbola in Point form
3. Equation of Normal of Hyperbola in Slope form
4. Solved Examples Based on Equation of Normal of Hyperbola

In this article, we will cover the concept of Equation of Normal in Point Form and Parametric Form. 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 fifteen questions have been asked on JEE MAINS( 2013 to 2023) from this topic including two in 2018, one in 2020, one in 2021, four in 2022, and one in 2023.

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 $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ where $b^2=a^2(e^2-1)$

Equation of Normal of Hyperbola in Point form

The equation of normal at $(x_1,y_1)$ 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 $({\mathrm{x}}_1,{\mathrm{y}}_1)$

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

$(\mathrm{y}-{\mathrm{y}}_1)=-\frac{{\mathrm{a}}^2{\mathrm{y}}_1}{{\mathrm{b}}^2{\mathrm{x}}_1}(\mathrm{x}-{\mathrm{x}}_1)$
$\text{ or }\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$
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 $ax\cos \theta +by\cot \theta =a^2+b^2$

Derivation of Equation of Normal of Hyperbola in Parametric form

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

$\begin{array}{rl}& \text{ Put }(x_1,y_1)\equiv (a\sec \theta ,b\tan \theta )\\ & \Rightarrow \frac{a^{2}x}{a\sec \theta }+\frac{b^{2}y}{b\tan \theta }=a^2+b^2\\ & \Rightarrow ax\cos \theta +by\cot \theta =a^2+b^2\end{array}$
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 $ax\cos \theta +by\cot \theta =a^2+b^2$

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 $(x_1,y_1)\equiv (a\sec \theta ,b\tan \theta )$.

$\begin{array}{rl}& \Rightarrow \frac{a^{2}x}{a\sec \theta }+\frac{b^{2}y}{b\tan \theta }=a^2+b^2\\ & \Rightarrow ax\cos \theta +by\cot \theta =a^2+b^2\end{array}$

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 $y=mx\mp \frac{m(a^2+b^2)}{\sqrt{a^2-m^2{b}^2}}$ and coordinate of point of contact is $(\pm \frac{a^2}{\sqrt{a^2-m^2{b}^2}},\mp \frac{m{b}^2}{\sqrt{a^2-m^2{b}^2}})$

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 $ax\cos \theta +by\cot \theta =a^2+b^2$

Derivation of Equation of Normal of Hyperbola in Slope form

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

Hence, the equation of normal becomes

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

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

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

$\begin{array}{rl}& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1\text{ is }(\frac{x^2}{a^2}-\frac{y^2}{b^2}-1)(\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}-1)={(\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}-1)}^2\\ & \text{ or, }S{S}_1=T^2\\ & \text{ where, }\mathrm{S}=\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}-1\\ & {\mathrm{S}}_1=\frac{{\mathrm{x}}_1^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}_1^2}{{\mathrm{b}}_2^2}-1\\ & \mathrm{T}=\frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}-\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2}-1\end{array}$
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 $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ 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 $QR$.

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 $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$
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 $\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1$
$y=mx\pm \sqrt{a^2-b^2{\mathrm{m}}^2}$
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 $1\mathrm{\&}\frac{3}{4}$
$\begin{array}{c}4y=3x-16\\ y=x-3\end{array}\}$ with positive intercept on $x$-axis.

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

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

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

Solution

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

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

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

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

tangent meet $\underset{―}{x}$-axis at $Q(1,0)$
latus rectum $x=\sqrt{6}$ at $R(\sqrt{6},\frac{2}{\sqrt{6}}(\sqrt{6}-1))$
Area of $\mathrm{\Delta }\mathrm{QFR}=\frac{1}{2}(\sqrt{6}-1)\cdot \frac{2}{\sqrt{6}}(\sqrt{6}-1)=\frac{(\sqrt{6}-1{)}^2}{\sqrt{6}}=\frac{7}{\sqrt{6}}-2$
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 $6x\cos \theta +3y\cot \theta =45$

$\begin{array}{rl}& M=-2\sin \theta =-\sqrt{2}\\ & \theta =\pi /4\end{array}$
The equation of normal is $\sqrt{2}x+y=15$

$\begin{array}{rl}& \mathrm{P}(\mathrm{asec}\theta ,b\tan \theta )\\ & \mathrm{P}(6\sqrt{2},3),\mathrm{k}(0,15)\\ & {\mathrm{d}}^2=216\end{array}$

Hence, the answer is 216