Ellipse
Ellipse for JEE Main & Advanced
Definition and Standard Form
Focal RadiiTopic 1
If $P(x_1, y_1)$ is any point lying on the boundary of the standard ellipse, its focal distances satisfy: \begin{formulabox}[Focal Distances] \[ SP = a - ex_1 \qquad (\text{from focus } S = (ae,0)) \] \[ S'P = a + ex_1 \qquad (\text{from focus } S'= (-ae,0)) \] \[ SP + S'P = 2a \quad (\text{constant value equal to major axis length}) \] \end{formulabox}
Vertical Ellipse: $b > a$Topic 2
When $b > a$ in $\ds\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the major axis shifts to lie vertically along the $y$-axis. The corresponding eccentricity balance matches $a^2 = b^2(1-e^2)$, the foci are situated at $(0, \pm be)$, and the horizontal directrices are $y = \pm \frac{b}{e}$.
Auxiliary Circle and Eccentric Angle
Eccentric Angle $\theta$Topic 1
Let $P$ be any point on the ellipse. Draw a vertical line through $P$ perpendicular to the major axis meeting it at $N$. Extend this line upward to meet the auxiliary circle at $Q$. The angle $\theta$ made by the radius vector $CQ$ with the positive direction of the x-axis is defined as the eccentric angle of $P$.
\begin{formulabox}[Parametric Equations] \[ x = a\cos\theta, \quad y = b\sin\theta, \quad \theta \in [0, 2\pi) \] \end{formulabox}
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- Ordinate Ratio Rule: The ratio of the vertical ordinate of the ellipse to that of the auxiliary circle at the same abscissa is constant: $\ds\frac{NP}{NQ} = \frac{b}{a}$.
- The eccentric angle $\theta$ is a purely geometric mapping parameter and does not equal the actual polar angle subtended by $P$ at the center $C$.
Equation of Tangent
Tangent at $(x_1, y_1)$ — Point FormTopic 1
\begin{formulabox}[Point Form ($T = 0$)] \[ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1 \] Derived using the standard $T = 0$ transformation rule: replace $x^2 \to x x_1$ and $y^2 \to y y_1$. \end{formulabox}
Tangent at Parameter $\theta$ — Parametric FormTopic 2
\begin{formulabox}[Parametric Form] \[ \frac{x\cos\theta}{a} + \frac{y\sin\theta}{b} = 1 \] \end{formulabox}
Tangent in Slope FormTopic 3
\begin{formulabox}[Slope Form — slope $m$] \[ y = mx \pm \sqrt{a^2 m^2 + b^2} \] The precise coordinate point of tangency contact maps to: $\left(\mp\dfrac{a^2 m}{\sqrt{a^2m^2+b^2}},\; \pm\dfrac{b^2}{\sqrt{a^2m^2+b^2}}\right)$. \end{formulabox}
\begin{importantbox}[Condition of Tangency] The straight line $y = mx + c$ is tangent to the standard ellipse $\ds\frac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ if and only if: \[ c^2 = a^2m^2 + b^2 \implies c = \pm \sqrt{a^2m^2+b^2}. \] \end{importantbox}
Properties of TangentsTopic 4
- Product of Perpendiculars: The product of the perpendicular distances dropped from the two foci $S$ and $S'$ to any variable tangent line is constant and always equal to $b^2$.
- The tangent line and the normal line at any point $P$ bisect the external and internal angles between the focal radii vectors $SP$ and $S'P$.
Equation of Normal
Normal at $(x_1, y_1)$ — Point FormTopic 1
\begin{formulabox}[Point Form] \[ \frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2 \] \end{formulabox}
Normal at Parameter $\theta$ — Parametric FormTopic 2
\begin{formulabox}[Parametric Form] \[ ax\sec\theta - by\csc\theta = a^2 - b^2 \] \end{formulabox}
Normal in Slope FormTopic 3
\begin{formulabox}[Slope Form — slope $m$] \[ y = mx \mp \frac{(a^2-b^2)m}{\sqrt{a^2 + b^2m^2}} \] \end{formulabox}
Number of Normals from an External PointTopic 4
From a general external point $P(h, k)$ lying in the coordinate plane, a maximum of four real normals can be drawn to an ellipse. Converting the slope equation format into an algebraic polynomial in terms of $m$ yields a degree-4 equation.
Co-normal PointsTopic 5
If the four normals drawn at the parameter points $\theta_1, \theta_2, \theta_3,$ and $\theta_4$ are concurrent at a single point, these boundary locations are called co-normal points. They satisfy the following trigonometric properties: \[ \sum \tan\theta_i = \tan\theta_1 + \tan\theta_2 + \tan\theta_3 + \tan\theta_4 = 0. \] Furthermore, the sum of their eccentric angles satisfies: $\theta_1 + \theta_2 + \theta_3 + \theta_4 = (2n+1)\pi, \, n \in \mathbb{Z}$.
Chord of the Ellipse
Chord Joining Parameters $\theta_1$ and $\theta_2$Topic 1
The equation of the straight line chord connecting two distinct parameter points $P(\theta_1)$ and $Q(\theta_2)$ on an ellipse is given by the trigonometric formula: \[ \frac{x}{a}\cos\left(\frac{\theta_1+\theta_2}{2}\right) + \frac{y}{b}\sin\left(\frac{\theta_1+\theta_2}{2}\right) = \cos\left(\frac{\theta_1-\theta_2}{2}\right). \] If a chord passes through the focus $S(ae, 0)$, it is a focal chord. Applying the focus coordinates to the chord equation yields the parametric condition: $\ds\tan\left(\frac{\theta_1}{2}\right)\tan\left(\frac{\theta_2}{2}\right) = \frac{e-1}{e+1}$.
Chord of ContactTopic 2
The straight line connecting the two points of tangency where tangent lines drawn from an external point $P(x_1, y_1)$ touch the ellipse is the chord of contact: \begin{formulabox}[Chord of Contact: $T = 0$] \[ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1 \] \end{formulabox}
Chord with a Given MidpointTopic 3
If a chord line is bisected exactly at an interior point $M(h, k)$, its equation can be found using the $T = S_1$ formula: \begin{formulabox}[Midpoint Form: $T = S_1$] \[ \frac{hx}{a^2} + \frac{ky}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} \] The corresponding slope of this bisected chord resolves to: $m = -\ds\frac{b^2 h}{a^2 k}$. \end{formulabox}
Pole and Polar
Pole and PolarTopic 1
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\begin{formulabox}[Polar of $(x_1,y_1)$ w.r.t. Ellipse] \[ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1 \quad \Longleftrightarrow \quad T = 0 \] The point $(x_1, y_1)$ is defined as the pole, and the resulting line is its polar line. \end{formulabox}
\begin{property} Conjugate Properties of Focus and Directrix: The polar line of a focus point $S(ae, 0)$ is its corresponding vertical directrix line $x = \frac{a}{e}$. Conversely, the pole of any directrix line is its matching focus. \end{property}
Conjugate Diameters
Conjugate DiametersTopic 1
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\begin{formulabox}[Conjugate Diameters] Two diameters of an ellipse (straight lines passing through the center) are defined as conjugate diameters if each diameter bisects all chords drawn parallel to the other.
If the slopes of the two conjugate diameters are $m_1$ and $m_2$, they satisfy the constant relationship: \[ m_1 \cdot m_2 = -\frac{b^2}{a^2} \] \end{formulabox}
Properties of Conjugate Diameters:- Tangent line boundary tracking: The tangent lines drawn at the endpoints of any diameter are parallel to its conjugate diameter.
- The major axis and minor axis of an ellipse form a pair of self-conjugate diameters.
- Eccentric Angle Relationship: The eccentric angles of the endpoints of two conjugate semi-diameters differ by exactly $90^\circ$: $\theta_2 - \theta_1 = \pm \frac{\pi}{2}$.
- The total area of the bounding parallelogram formed by drawing tangent lines at the endpoints of a pair of conjugate diameters is constant and always equal to $4ab$.
- Apollonius Theorem: The sum of the squares of the lengths of any two conjugate semi-diameters $CP$ and $CD$ is constant and equal to the sum of the squares of the semi-axes: $CP^2 + CD^2 = a^2 + b^2$.
Director Circle
Director CircleTopic 1
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\begin{formulabox}[Director Circle of Ellipse] The locus of the point of intersection of two tangents to an ellipse that meet at a constant angle of $90^\circ$ is a concentric circle called the director circle. Its equation is: \[ x^2 + y^2 = a^2 + b^2 \] \end{formulabox}
For the standard ellipse $\ds\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the radius of the director circle is $\sqrt{a^2+b^2}$.
Subtangent and Subnormal
Subtangent and SubnormalTopic 1
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Let the tangent line at any boundary point $P(x_1, y_1)$ on an ellipse intersect the major axis at $T$, and let the normal line at $P$ intersect the major axis at $G$. If $N(x_1, 0)$ is the foot of the perpendicular dropped from $P$ onto the major axis, the lengths of the projections are defined as:
\begin{formulabox}[Subtangent and Subnormal Proportions] \[ \text{Length of Subtangent } (NT) = \left| \frac{a^2}{x_1} - x_1 \right| = \frac{a^2 - x_1^2}{|x_1|} \] \[ \text{Length of Subnormal } (NG) = \left| x_1 - \frac{b^2 x_1}{a^2} \right| = |x_1|\left(1 - \frac{b^2}{a^2}\right) = |x_1|e^2 \] \end{formulabox}
The length of the subnormal is directly proportional to the abscissa of the point and depends on the eccentricity of the ellipse.
Reflection Property of Ellipse
Reflection Property of EllipseTopic 1
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\begin{importantbox}[Reflection Property] An incoming ray of light originating from one focus of an ellipse passes through the other focus after reflecting off the inner curve boundary surface.
This is because the normal line at any point $P$ on the ellipse acts as the angle bisector of the focal angle $\angle SPS'$ formed by the focal radii vectors. \end{importantbox}
Special Results and JEE Tricks
Special Results and JEE TricksTopic 1
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\begin{warningbox}[JEE Important Results — Ellipse]- Focal Product Property: For any point $P(x_1, y_1)$ lying on the ellipse, the product of its focal distances satisfies: $SP \cdot S'P = a^2 - e^2 x_1^2$.
- The locus of the foot of the perpendicular dropped from either focus to any variable tangent line is the auxiliary circle $x^2 + y^2 = a^2$.
- The equation of the tangent lines drawn at the endpoints of the latus rectum can be simplified directly to the normal format: $\pm ex \pm \ds\frac{y}{a} = 1$.
- Symmetric Area Mapping: The area of any polygon inscribed inside an ellipse is related to the area of the corresponding polygon inscribed inside its auxiliary circle by the scaling ratio: $\ds\frac{\text{Area}_{\text{ellipse}}}{\text{Area}_{\text{circle}}} = \frac{b}{a}$.
Practice Problems (JEE Level)
Practice Problems (JEE Level)Topic 1
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- If the tangent line drawn at the upper endpoint of the latus rectum of the ellipse $\ds\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intersects the major axis at a point $T$, show that the distance from the origin to $T$ satisfies $OT = \frac{a}{e}$ (the location of the directrix line).
- Prove that the product of the lengths of the perpendicular distances dropped from the two foci $S$ and $S'$ to any variable tangent line of an ellipse is constant and equal to $b^2$.
- Find the equation of the chord of the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{9} = 1$ that is bisected exactly at the point $(2, 1)$.
- Find the mathematical condition on the line parameters for which the straight line $lx + my = 1$ is a valid tangent line to the standard ellipse $\ds\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
- Show that the sum of the squares of the lengths of any two conjugate semi-diameters of an ellipse is constant and equal to $a^2 + b^2$.
- If the line $lx + my = 1$ is tangent to the ellipse $\ds\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the coefficients must satisfy which condition?
- $a^2l^2 + b^2m^2 = 1$
- $a^2l^2 - b^2m^2 = 1$
- $a^2m^2 + b^2l^2 = 1$
- $l^2 + m^2 = a^2 + b^2$
- Find the distance from the origin to the intersection point $T$ of the tangent at the latus rectum endpoint on $\ds\frac{x^2}{16} + \frac{y^2}{12} = 1$ (where $e = 1/2$):
- $8$
- $4$
- $16$
- $2$
- Find the equation of the chord of $\ds\frac{x^2}{25} + \frac{y^2}{9} = 1$ bisected exactly at $(2, 1)$:
- $18x + 25y = 61$
- $9x + 25y = 43$
- $18x + 25y = 89$
- $2x + y = 5$
- For the ellipse $\ds\frac{x^2}{9} + \frac{y^2}{4} = 1$, the sum of the squared lengths of any two conjugate semi-diameters $CP^2 + CD^2$ evaluates to:
- $13$
- $5$
- $25$
- $9$
- A tangent to $\ds\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ forms a right-angled triangle with the coordinate axes. The minimum possible area of this triangle is:
- $ab$
- $2ab$
- $\frac{1}{2}ab$
- $a^2 + b^2$
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Quick Reference Card
Quick Reference CardTopic 1
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\begin{formulabox}[All Ellipse Formulas at a Glance] Standard Equation Parameters: $\ds\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b>0$, $b^2=a^2(1-e^2)$.
\medskip
| Geometric Operational Element | Algebraic Formula / Line Equation Form |
| Foci Coordinates ($S, S'$): | $(\pm ae, 0)$ |
| Directrices Equations ($L, L'$): | $x = \pm a/e$ |
| Parametric Boundary Coordinates: | $(a\cos\theta, b\sin\theta)$ |
| Tangent Line Equation (Point Form): | $\ds\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$ |
| Tangent Line Equation (Slope Form): | $y = mx \pm \sqrt{a^2m^2+b^2}$ |
| Condition of Tangency: | $c^2 = a^2m^2+b^2$ |
| Normal Line Equation (Point Form): | $\ds\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$ |
| Chord of Contact Line Equation: | $\ds\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$ |
| Chord line with Midpoint $(h,k)$: | $T = S_1 \implies \ds\frac{hx}{a^2} + \frac{ky}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2}$ |
| Director Circle Equation: | $x^2+y^2=a^2+b^2$ |
| String Property Focus Vector Sum: | $SP+S'P=2a$ |
| Conjugate Diameters Slope Product: | $m_1 \cdot m_2 = -b^2/a^2$ |
Solved Examples
Worked ExamplesTopic 1
Example 1 — Tangent and Normal Formulations
Find the equations of the tangent line and the normal line to the ellipse $\ds\frac{x^2}{9} + \frac{y^2}{4} = 1$ at the parameter point parameterized at an eccentric angle $\theta = \frac{\pi}{3}$.
Show solution
- Tangent Line Equation (Parametric Form):
Substitute $\theta = \frac{\pi}{3}$ into the standard parametric tangent formula: \[ \frac{x\cos(\pi/3)}{3} + \frac{y\sin(\pi/3)}{2} = 1 \implies \frac{x(1/2)}{3} + \frac{y(\sqrt{3}/2)}{2} = 1 \implies \frac{x}{6} + \frac{\sqrt{3}y}{4} = 1. \] Multiply by the common denominator $12$ to simplify: $2x + 3\sqrt{3}y - 12 = 0$. - Normal Line Equation (Parametric Form):
Substitute the parameters into the parametric normal formula: \[ 3x\sec\left(\frac{\pi}{3}\right) - 2y\csc\left(\frac{\pi}{3}\right) = a^2 - b^2 \implies 3x(2) - 2y\left(\frac{2}{\sqrt{3}}\right) = 9 - 4 \] \[ 6x - \frac{4}{\sqrt{3}}y = 5 \implies 6\sqrt{3}x - 4y - 5\sqrt{3} = 0. \]
Example 2 — Director Circle Locus Tracking
Find the locus of the point of intersection of two mutually perpendicular tangent lines drawn to the ellipse $9x^2 + 16y^2 = 144$.
Show solution
First, normalize the ellipse equation by dividing all terms by $144$: \[ \frac{9x^2}{144} + \frac{16y^2}{144} = 1 \implies \frac{x^2}{16} + \frac{y^2}{9} = 1. \] This matches the standard form where $a^2 = 16$ and $b^2 = 9$. The locus of the point of intersection of perpendicular tangents is defined as the director circle. Apply its standard formula: \[ x^2 + y^2 = a^2 + b^2 \implies x^2 + y^2 = 16 + 9 \implies x^2 + y^2 = 25. \] Final Answer: The locus is the circle $x^2 + y^2 = 25$.
Example 3 — Chord calculations through Midpoint Form
Find the equation of the chord of the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{16} = 1$ that is bisected exactly at the interior point $M(2, 1)$.
Show solution
Identify the midpoint parameters: $h = 2$ and $k = 1$. Apply the $T = S_1$ midpoint formula: \[ \frac{hx}{a^2} + \frac{ky}{b^2} = \frac{h^2}{a^2} + \frac{k^2}{b^2} \implies \frac{2x}{25} + \frac{1y}{16} = \frac{2^2}{25} + \frac{1^2}{16} \] \[ \frac{2x}{25} + \frac{y}{16} = \frac{4}{25} + \frac{1}{16} \implies \frac{32x + 25y}{400} = \frac{64 + 25}{400} \implies 32x + 25y = 89. \] Final Answer: The chord equation is $32x + 25y - 89 = 0$.