Length of Sub-Tangent and Sub-Normal of an Ellipse

What are Sub-Tangent and Sub-Normal of an Ellipse?

The concepts of sub-tangent and sub-normal arise when a tangent and a normal are drawn to an ellipse at a given point. These lengths help mathematicians study the geometry of curves and understand how the tangent and normal interact with the coordinate axes. They are important topics in coordinate geometry, differential calculus, and conic sections.

Sub-Tangent Meaning in Simple Words

The sub-tangent is the portion of the x-axis intercepted between the foot of the ordinate drawn from a point on the ellipse and the point where the tangent meets the x-axis.

In simple terms, it measures how far the tangent extends along the x-axis from the point directly below the point of contact.

Sub-Normal Meaning in Simple Words

The sub-normal is the portion of the x-axis intercepted between the foot of the ordinate and the point where the normal meets the x-axis.

It represents the horizontal distance associated with the normal line drawn at a point on the ellipse.

Definition of Sub-Tangent

The sub-tangent of a curve at a point is the segment of the x-axis between the foot of the ordinate and the point where the tangent intersects the x-axis.

If the tangent at $(x,y)$ has slope $\frac{dy}{dx}$, then

$\text{Sub-Tangent}=\frac{y}{\frac{dy}{dx}}$

Definition of Sub-Normal

The sub-normal of a curve at a point is the segment of the x-axis between the foot of the ordinate and the point where the normal intersects the x-axis.

It is given by

$\text{Sub-Normal}=y\frac{dy}{dx}$

Why Sub-Tangent and Sub-Normal are Important

Sub-tangent and sub-normal provide useful information about the behavior of curves and the geometry of tangents and normals. They are widely used in:

• Coordinate geometry

• Differential calculus

• Conic sections

• Engineering design

• Optics and reflection problems

• Curve tracing and analysis

Basics of Ellipse Geometry

Before studying sub-tangent and sub-normal formulas, it is important to understand the basic properties of an ellipse.

What is an Ellipse?

An ellipse is a closed conic section obtained when a plane cuts a cone at an angle smaller than the angle of the cone's side.

It is also defined as the locus of a point whose sum of distances from two fixed points called foci remains constant.

Examples of elliptical shapes include:

• Planetary orbits

• Running tracks

• Satellite trajectories

• Reflective mirrors

Standard Equation of an Ellipse

The standard equation of an ellipse centered at the origin is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

where:

• $a$ is the semi-major axis

• $b$ is the semi-minor axis

• $a>b$

If the major axis is vertical, the equation becomes

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$

Major Axis and Minor Axis

The longest diameter of an ellipse is called the major axis.

Its length is $2a$

The shortest diameter is called the minor axis.

Its length is $2b$

These axes determine the overall shape and size of the ellipse.

Coordinates of a Point on an Ellipse

A point on the ellipse can be represented parametrically as

$(a\cos\theta,b\sin\theta)$

where $\theta$ is called the eccentric angle or parameter.

This form simplifies many calculations involving tangents, normals, sub-tangents, and sub-normals.

Tangent and Normal to an Ellipse

Tangents and normals are fundamental geometric concepts used to study curves.

Equation of Tangent to an Ellipse

The equation of the tangent at the point $(x_1,y_1)$ on

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$

This equation is widely used in coordinate geometry problems.

Equation of Normal to an Ellipse

The equation of the normal at $(x_1,y_1)$ is

$\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$

The normal is always perpendicular to the tangent at the point of contact.

Slope of Tangent

Differentiating

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

gives

$\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$

Hence,

$\frac{dy}{dx}=-\frac{b^2x}{a^2y}$

This is the slope of the tangent.

Slope of Normal

Since the normal is perpendicular to the tangent,

$\text{Slope of Normal}=-\frac{1}{\text{Slope of Tangent}}$

Therefore,

$m_n=\frac{a^2y}{b^2x}$

Length of Sub-Tangent of an Ellipse

The sub-tangent provides a measure of how the tangent intersects the x-axis.

Formula for Length of Sub-Tangent

The general formula is

$\text{Sub-Tangent}=\frac{y}{\frac{dy}{dx}}$

For the ellipse,

$\frac{dy}{dx}=-\frac{b^2x}{a^2y}$

Therefore,

$\text{Sub-Tangent}=-\frac{a^2y^2}{b^2x}$

Its magnitude is

$\left|\frac{a^2y^2}{b^2x}\right|$

Derivation of the Sub-Tangent Formula

Using the tangent slope,

$\text{Sub-Tangent}=\frac{y}{\frac{dy}{dx}}$

Substituting

$\frac{dy}{dx}=-\frac{b^2x}{a^2y}$

gives

$\text{Sub-Tangent}=-\frac{a^2y^2}{b^2x}$

This formula directly relates the coordinates of the point to the length of the sub-tangent.

Parametric Form of Sub-Tangent

Using

$x=a\cos\theta,\quad y=b\sin\theta$

we obtain

$\text{Sub-Tangent}=\frac{a\sin^2\theta}{\cos\theta}$

This form is useful in advanced coordinate geometry problems.

Geometric Interpretation of Sub-Tangent

Geometrically, the sub-tangent represents the horizontal projection associated with the tangent line.

Its value changes continuously as the point moves along the ellipse.

Length of Sub-Normal of an Ellipse

The sub-normal measures a corresponding horizontal distance related to the normal.

Formula for Length of Sub-Normal

The general formula is

$\text{Sub-Normal}=y\frac{dy}{dx}$

For the ellipse,

$\text{Sub-Normal}=y\left(-\frac{b^2x}{a^2y}\right)$

Hence,

$\text{Sub-Normal}=-\frac{b^2x}{a^2}$

Its magnitude is

$\left|\frac{b^2x}{a^2}\right|$

Derivation of the Sub-Normal Formula

Starting with

$\text{Sub-Normal}=y\frac{dy}{dx}$

and substituting $\frac{dy}{dx}=-\frac{b^2x}{a^2y}$ gives $\text{Sub-Normal}=-\frac{b^2x}{a^2}$

Parametric Form of Sub-Normal

Using $x=a\cos\theta$ the formula becomes

$\text{Sub-Normal}=-\frac{b^2\cos\theta}{a}$

Geometric Interpretation of Sub-Normal

The sub-normal measures the horizontal displacement associated with the normal line and provides information about the curvature of the ellipse.

Derivation of Sub-Tangent and Sub-Normal Formulas

The formulas can be obtained through several mathematical approaches.

Using Differential Calculus

Differentiating the ellipse equation gives the tangent slope. Substituting this slope into the standard formulas yields the sub-tangent and sub-normal.

Using Slope of the Tangent

Since both quantities depend on $\frac{dy}{dx}$, knowledge of the tangent slope is sufficient to derive them.

Coordinate Geometry Approach

Using the equations of the tangent and normal, their intersections with the x-axis can be determined directly, leading to the same formulas.

Parametric Derivation

Replacing coordinates with $x=a\cos\theta$ and $y=b\sin\theta$

provides elegant parametric expressions often used in advanced geometry.

Properties of Sub-Tangent and Sub-Normal

Sub-tangent and sub-normal possess several useful properties.

Dependence on the Point of Contact

Their values depend on the coordinates of the point where the tangent and normal are drawn.

Different points produce different lengths.

Relationship with Tangent and Normal

The sub-tangent is associated with the tangent line, while the sub-normal is associated with the normal line.

Both are derived from the slope of the tangent.

Symmetry Properties of the Ellipse

Because an ellipse is symmetric about both axes, sub-tangent and sub-normal values exhibit corresponding symmetry.

Important Geometric Observations

• Both quantities vary continuously.

• They depend on the curvature of the ellipse.

• Their values become simpler in parametric form.

Sub-Tangent and Sub-Normal in Parametric Form

Parametric equations simplify calculations involving conic sections.

Parametric Coordinates of an Ellipse

A point on the ellipse is represented by

$(a\cos\theta,b\sin\theta)$

This form avoids complicated algebraic manipulations.

Sub-Tangent in Terms of $\theta$

Using parametric coordinates,

$\text{Sub-Tangent}=\frac{a\sin^2\theta}{\cos\theta}$

Sub-Normal in Terms of $\theta$

The corresponding parametric expression is

$\text{Sub-Normal}=-\frac{b^2\cos\theta}{a}$

Advantages of Parametric Representation

• Simplifies derivations.

• Reduces algebraic complexity.

• Useful in JEE and advanced coordinate geometry problems.

• Provides compact formulas.

Applications of Sub-Tangent and Sub-Normal

These concepts have several theoretical and practical applications.

Applications in Coordinate Geometry

Used for solving tangent-normal problems involving conic sections.

Applications in Differential Calculus

Help analyze the local behavior of curves using derivatives.

Applications in Optics and Engineering

Used in reflector design, optical systems, and mechanical structures involving elliptical shapes.

Applications in Curve Analysis

Assist in studying curvature, slope variation, and geometric properties of curves.

Difference Between Sub-Tangent and Sub-Normal

Although related, they represent different geometric quantities.

Definition Comparison

Sub-tangent is associated with the tangent line.

Sub-normal is associated with the normal line.

Formula Comparison

Sub-tangent:

$\frac{y}{\frac{dy}{dx}}$

Sub-normal:

$y\frac{dy}{dx}$

Geometrical Difference

Sub-tangent measures a distance related to the tangent.

Sub-normal measures a distance related to the normal.

Comparison Table

Important Results Related to Ellipse Tangents and Normals

Several standard formulas are repeatedly used in ellipse problems.

Standard Tangent Equations

At $(x_1,y_1)$,

$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$

Parametric tangent:

$\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1$

Standard Normal Equations

At $(x_1,y_1)$,

$\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$

Length Formulas Summary

Sub-Tangent:

$\text{ST}=\frac{y}{dy/dx}$

Sub-Normal:

$\text{SN}=y\frac{dy}{dx}$

For the ellipse,

$\text{ST}=-\frac{a^2y^2}{b^2x}$

$\text{SN}=-\frac{b^2x}{a^2}$

These formulas form the foundation of most problems involving the length of sub-tangent and sub-normal of an ellipse.

Best Books for Length of Sub-Tangent and Sub-Normal of an Ellipse

The concepts of sub-tangent and sub-normal are important applications of coordinate geometry and conic sections. The following books provide detailed explanations and solved examples.

Shortcut Tips and Tricks for Length of Sub-Tangent and Sub-Normal of an Ellipse

A few geometric observations can simplify calculations involving tangents and normals to an ellipse.

Important Formula Table

The following formulas are frequently used while finding the lengths of the sub-tangent and sub-normal of an ellipse.

Solved Examples based on the Length of the Subtangent and Subnormal of Ellipse

Example 1: If the normal to the ellipse $3x^2+4y^2=12$ at a point $P$ on it is parallel to the line $2x+y=4$ and the tangent to the ellipse at $P$ passes through $Q(4,4)$, then $PQ$ is equal to:

1. $\frac{\sqrt{157}}{2}$
2. $\frac{\sqrt{221}}{2}$
3. $\frac{\sqrt{61}}{2}$
4. $\frac{5\sqrt{5}}{2}$

Solution:

Equation of normal to the ellipse:

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

$\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$

Distance formula:

The distance between the points $A(x_1,y_1)$ and $B(x_2,y_2)$ is

$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Given,

$\begin{aligned}
3x^2+4y^2&=12\
\frac{x^2}{4}+\frac{y^2}{3}&=1
\end{aligned}$

Therefore,

$x=2\cos\theta,\qquad y=\sqrt{3}\sin\theta$

Let

$P(2\cos\theta,\sqrt{3}\sin\theta)$

Equation of normal:

$\frac{4x}{2\cos\theta}-\frac{3y}{\sqrt{3}\sin\theta}=4-3$

$\Rightarrow 2x\sin\theta-\sqrt{3}y\cos\theta=\cos\theta\sin\theta$

Slope of normal

$\frac{2}{\sqrt{3}}\tan\theta=-2$

$\Rightarrow \tan\theta=-\sqrt{3}$

Since the tangent passes through $(4,4)$,

$12\cos\theta+8\sqrt{3}\sin\theta=6$

$\frac{1}{2}\cos\theta+\frac{8\sqrt{3}}{6}\sin\theta=1$

$\cos\theta=-\frac{1}{2},\qquad \sin\theta=\frac{\sqrt{3}}{2}$

$\therefore \theta=120^\circ$

Hence,

$P=(2\cos120^\circ,\sqrt{3}\sin120^\circ)$

$\begin{aligned}
P\left(-1,\frac{3}{2}\right),\qquad Q(4,4)
\end{aligned}$

$\begin{aligned}
PQ&=\sqrt{(-1-4)^2+\left(\frac{3}{2}-4\right)^2}\
&=\sqrt{25+\frac{25}{4}}\
&=\frac{5\sqrt{5}}{2}
\end{aligned}$

Hence, the answer is option (4).

Example 2: Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x^2}{2}+\frac{y^2}{4}=1$ meet at the point $R(\sqrt{2},2\sqrt{2}-2)$. If $S$ is the focus of the ellipse on its negative major axis, then $SP^2+SQ^2$ is equal to:

Solution:

Equation of chord of contact is $T=0$.

$\frac{\sqrt{2}x}{2}+\frac{(2\sqrt{2}-2)y}{4}=1$

$\Rightarrow 2\sqrt{2}x+(2\sqrt{2}-2)y=4$

$\Rightarrow x=\frac{4-(2\sqrt{2}-2)y}{2\sqrt{2}}$

Finding the point of intersection with the ellipse:

$\frac{(4-(2\sqrt{2}-2)y)^2}{16}+\frac{y^2}{4}=1$

$\Rightarrow \frac{(2-(\sqrt{2}-1)y)^2}{4}+\frac{y^2}{4}=1$

$\Rightarrow 4+(\sqrt{2}-1)^2y^2-4(\sqrt{2}-1)y+y^2=4$

$\Rightarrow (3-2\sqrt{2}+1)y^2-4(\sqrt{2}-1)y=0$

$\Rightarrow 2\sqrt{2}(\sqrt{2}-1)y^2-4(\sqrt{2}-1)y=0$

$\Rightarrow 2\sqrt{2}(\sqrt{2}-1)y(y-\sqrt{2})=0$

$\Rightarrow y=0,\sqrt{2}$

Corresponding values of $x$ are

$x=\sqrt{2},1$

Therefore,

$P(\sqrt{2},0),\qquad Q(1,\sqrt{2})$

For the ellipse,

$e=\sqrt{1-\frac{2}{4}}=\frac{1}{\sqrt{2}}$

Hence,

$S=(0,-\sqrt{2})$

$\begin{aligned}
SP^2&=(\sqrt{2}-0)^2+(0+\sqrt{2})^2\
&=2+2=4
\end{aligned}$

$\begin{aligned}
SQ^2&=(1-0)^2+(\sqrt{2}+\sqrt{2})^2\
&=1+8=9
\end{aligned}$

Therefore,

$SP^2+SQ^2=4+9=13$

Hence, the answer is 13.

Example 3: If the normal at the point $P(\theta)$ to the ellipse $\frac{x^2}{14}+\frac{y^2}{5}=1$ intersects it again at the point $Q(2\theta)$, then $\cos\theta$ is equal to:

1. $\frac{2}{3}$
2. $-\frac{2}{3}$
3. $\frac{3}{4}$
4. None of these

Solution:

The equation of the normal is

$\frac{\sqrt{14}x}{\cos\theta}-\frac{\sqrt{5}y}{\sin\theta}=14-5$

Since it passes through

$(\sqrt{14}\cos2\theta,\sqrt{5}\sin2\theta)$

we get

$\frac{14(2\cos^2\theta-1)}{\cos\theta}-\frac{5(2\sin\theta\cos\theta)}{\sin\theta}=9$

$\Rightarrow 28\cos\theta-\frac{14}{\cos\theta}-10\cos\theta=9$

$\Rightarrow 18\cos^2\theta-9\cos\theta-14=0$

$\Rightarrow (3\cos\theta+2)(6\cos\theta-7)=0$

$\Rightarrow \cos\theta=-\frac{2}{3}$

Hence, the answer is option (2).

Example 4: Let the tangent and normal at the point $(3\sqrt{3},1)$ on the ellipse $\frac{x^2}{36}+\frac{y^2}{4}=1$ meet the $y$-axis at the points $A$ and $B$ respectively. Let the circle $C$ be drawn taking $AB$ as a diameter and the line $x=2\sqrt{5}$ intersect $C$ at the points $P$ and $Q$. If the tangents at the points $P$ and $Q$ on the circle intersect at the point $(\alpha,\beta)$, then $(\alpha^2-\beta^2)$ is equal to: [JEE Main 2023]

Solution:

Given ellipse

$\frac{x^2}{36}+\frac{y^2}{4}=1$

The tangent at $(3\sqrt{3},1)$ is

$\frac{x(3\sqrt{3})}{36}+\frac{y(1)}{4}=1$

$\Rightarrow \frac{x}{4\sqrt{3}}+\frac{y}{4}=1$

Putting $x=0$, we get

$y=4$

Therefore,

$A=(0,4)$

The normal at $(3\sqrt{3},1)$ is

$\frac{x}{4}-\frac{4y}{4\sqrt{3}}=\frac{2}{\sqrt{3}}$

Putting $x=0$, we get

$y=-8$

Therefore,

$B=(0,-8)$

The circle having $AB$ as diameter has center

$(0,-2)$

and radius

$6$

Hence, its equation is

$x^2+(y+2)^2=36$

Let the point of intersection of tangents at $P$ and $Q$ be $(h,k)$.

Using the chord of contact relation,

$hx+ky+2(y+k)-32=0$

Comparing coefficients,

$k=-2$

Thus,

$hx+2k-32=0$

$\Rightarrow hx=36$

Since $x=2\sqrt{5}$,

$h=\frac{36}{2\sqrt{5}}$

Therefore,

$\alpha=\frac{36}{2\sqrt{5}}$

and

$\beta=-2$

Hence,

$\begin{aligned}
\alpha^2-\beta^2
&=\left(\frac{36}{2\sqrt{5}}\right)^2-(-2)^2\
&=\frac{324}{5}-4\
&=\frac{324-20}{5}\
&=\frac{304}{5}
\end{aligned}$

Hence, the answer is $\frac{304}{5}$.

Example 5: If $m$ is the slope of a common tangent to the curves $\frac{x^2}{16}+\frac{y^2}{9}=1$ and $x^2+y^2=12$, then $12m^2$ is equal to: [JEE Main 2022]

Solution:

Given ellipse

$\frac{x^2}{16}+\frac{y^2}{9}=1$

The equation of a tangent with slope $m$ is

$y=mx\pm\sqrt{16m^2+9}$

For the circle

$x^2+y^2=12$

the equation of a tangent with slope $m$ is

$y=mx\pm\sqrt{12m^2+12}$

Since the tangent is common to both curves,

$\sqrt{16m^2+9}=\sqrt{12m^2+12}$

Squaring both sides,

$16m^2+9=12m^2+12$

$\Rightarrow 4m^2=3$

$\Rightarrow m^2=\frac{3}{4}$

Therefore,

$\begin{aligned}
12m^2
&=12\left(\frac{3}{4}\right)\
&=9
\end{aligned}$

Hence, the answer is 9.

Related Topics to Length of Sub-Tangent and Sub-Normal of an Ellipse

The concepts of sub-tangent and sub-normal are closely connected with ellipses, tangents, normals, parametric equations, differentiation, and coordinate geometry. Exploring these topics helps develop a stronger understanding of conic sections and curve analysis.