JEE Main & Advanced

Ellipse

Ellipse for JEE Main & Advanced

1
Module 1

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 $y = \pm \frac{b}{e}$.

✎ Self-Check — 5 questions0 / 5
Q1.Find the eccentricity of the ellipse represented by the equation $9x^2 + 25y^2 = 225$:
Q2.A variable point moves such that the sum of its distances from two fixed points $S(3,0)$ and $S'(-3,0)$ is constantly equal to $10$ units. The equation of its locus profile matches:
Q3.Find the total length of the latus rectum for the vertical ellipse configuration given by $16x^2 + 9y^2 = 144$:
Q4.Find the distance separating the two parallel vertical directrices of the ellipse $\ds\frac{x^2}{36} + \frac{y^2}{20} = 1$:
Q5.If a point $P$ lying on the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{7} = 1$ has its abscissa coordinate equal to $x_1 = 2$, calculate its focal distance $SP$ from the positive focus:
2
Module 2

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}

    [leftmargin=*]
  • 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 $C$.
✎ Self-Check — 5 questions0 / 5
Q1.Find the eccentric angle $\theta$ of a point lying on the ellipse $\ds\frac{x^2}{4} + \frac{y^2}{1} = 1$ whose coordinates are given by $\left(1, \frac{\sqrt{3}}{2}\right)$:
Q2.If the vertical ordinate of a point $P$ on an ellipse is $y_P = 3$, and the corresponding ordinate of point $Q$ on its auxiliary circle is $y_Q = 5$, find the ratio of the semi-minor axis to the semi-major axis ($\frac{b}{a}$):
Q3.Find the standard radius of the auxiliary circle associated with the shifted ellipse equation $\ds\frac{(x-2)^2}{16} + \frac{(y+3)^2}{9} = 1$:
Q4.A parameter point on $\ds\frac{x^2}{100} + \frac{y^2}{64} = 1$ is designated at an eccentric angle $\theta = \frac{\pi}{4}$. Its exact coordinate position matches:
Q5.Find the equation of the auxiliary circle for the vertical ellipse configuration $9x^2 + 4y^2 = 36$:
3
Module 3

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

  1. 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 $b^2$.
  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 $S'P$.
✎ Self-Check — 5 questions0 / 5
Q1.Find the equation of the tangent line to the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{9} = 1$ that has a slope of $m = 1$ and a positive y-intercept:
Q2.Find the equation of the tangent to the ellipse $4x^2 + 9y^2 = 36$ at the parameter point whose eccentric angle is $\theta = \frac{\pi}{4}$:
Q3.For what value of the constant intercept parameter $c$ is the straight line $y = 2x + c$ tangent to the ellipse $\ds\frac{x^2}{4} + \frac{y^2}{1} = 1$?
Q4.Find the coordinate position where the tangent line $\ds\frac{x}{4} + \frac{y}{3} = \sqrt{2}$ touches the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{9} = 1$:
Q5.Calculate the product of the perpendicular distances dropped from the foci of the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{16} = 1$ to the tangent line $y = x + \sqrt{41}$:
4
Module 4

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: $\theta_1 + \theta_2 + \theta_3 + \theta_4 = (2n+1)\pi, \, n \in \mathbb{Z}$.

✎ Self-Check — 5 questions0 / 5
Q1.Find the equation of the normal line to the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{9} = 1$ at the parameter point whose eccentric angle is $\theta = \frac{\pi}{3}$:
Q2.Find the equation of the normal line to the ellipse $4x^2 + 9y^2 = 36$ at the specific point coordinate position $\left(\frac{3}{2}, \sqrt{3}\right)$:
Q3.The maximum number of real normal lines that can be drawn from an external point to an ellipse is:
Q4.If the eccentric angles of four co-normal points on an ellipse are $\theta_1, \theta_2, \theta_3,$ and $\theta_4$, the value of $\sum \tan\theta_i$ is identically:
Q5.Find the slope of the normal line to the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{16} = 1$ at the point $(0, 4)$:
5
Module 5

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}

✎ Self-Check — 5 questions0 / 5
Q1.Find the equation of the chord of the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{9} = 1$ whose midpoint is located at $M(2, 1)$:
Q2.Find the slope of a chord of the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{4} = 1$ that is bisected exactly at the point $(1, -1)$:
Q3.Find the equation of the chord of contact drawn from the external point $P(5, 4)$ to the ellipse $4x^2 + 9y^2 = 36$:
Q4.If the parameter endpoints of a focal chord passing through the positive focus of an ellipse are $\theta_1$ and $\theta_2$, the product $\ds\tan\left(\frac{\theta_1}{2}\right)\tan\left(\frac{\theta_2}{2}\right)$ must equal:
Q5.Find the equation of the chord connecting the two points on $\ds\frac{x^2}{4} + \frac{y^2}{1} = 1$ whose eccentric angles are $\theta_1 = \frac{\pi}{2}$ and $\theta_2 = 0$:
6
Module 6

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}

✎ Self-Check — 5 questions0 / 5
Q1.Find the polar line of the point $P(2, 3)$ with respect to the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{9} = 1$:
Q2.Find the pole coordinates of the straight line $x + y = 2$ with respect to the ellipse $\ds\frac{x^2}{8} + \frac{y^2}{4} = 1$:
Q3.The polar line of the negative focus $S'(-ae, 0)$ with respect to a standard ellipse is:
Q4.If the points $(2, 1)$ and $(4, k)$ are conjugate points with respect to the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{4} = 1$, find the value of $k$:
Q5.The locus of the pole of a variable line that always stands at a constant perpendicular distance from the center of an ellipse forms a concentric:
7
Module 7

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:
  1. Tangent line boundary tracking: The tangent lines drawn at the endpoints of any diameter are parallel to its conjugate diameter.
  2. The major axis and minor axis of an ellipse form a pair of self-conjugate diameters.
  3. Eccentric Angle Relationship: The eccentric angles of the endpoints of two conjugate semi-diameters differ by exactly $90^\circ$: $\theta_2 - \theta_1 = \pm \frac{\pi}{2}$.
  4. 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 $4ab$.
  5. 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: $CP^2 + CD^2 = a^2 + b^2$.
✎ Self-Check — 5 questions0 / 5
Q1.If one diameter of the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{9} = 1$ has a slope $m_1 = 3$, find the slope $m_2$ of its conjugate diameter:
Q2.The eccentric angles of the endpoints of two conjugate semi-diameters on an ellipse must differ by exactly:
Q3.Calculate the area of the bounding parallelogram formed by drawing tangent lines at the endpoints of the conjugate diameters of the ellipse $9x^2 + 16y^2 = 144$:
Q4.For any pair of conjugate semi-diameters $CP$ and $CD$ on the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{16} = 1$, the sum of their squared lengths $CP^2 + CD^2$ evaluates to:
Q5.If the slopes $m_1$ and $m_2$ of two conjugate diameters of an ellipse are equal in magnitude but opposite in sign ($m_2 = -m_1$), the diameters are called equi-conjugate. Their common slope magnitude satisfies:
8
Module 8

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 $\sqrt{a^2+b^2}$.

✎ Self-Check — 5 questions0 / 5
Q1.Find the equation of the director circle for the ellipse represented by $9x^2 + 16y^2 = 144$:
Q2.Find the radius of the director circle associated with the ellipse $\ds\frac{x^2}{36} + \frac{y^2}{28} = 1$:
Q3.Perpendicular tangents are drawn to the ellipse $\ds\frac{x^2}{10} + \frac{y^2}{6} = 1$ from a point $P$. The distance from the origin to $P$ is constant and equal to:
Q4.Find the equation of the director circle for the shifted ellipse configuration given by $\ds\frac{(x-1)^2}{9} + \frac{(y-2)^2}{4} = 1$:
Q5.If the director circle of the ellipse $\ds\frac{x^2}{a^2} + \frac{y^2}{9} = 1$ passes through the point $(4, 3)$, find the value of the semi-major axis $a$:
9
Module 9

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.

✎ Self-Check — 5 questions0 / 5
Q1.Find the length of the subnormal to the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{16} = 1$ at a boundary point whose abscissa is $x_1 = 5$:
Q2.Find the length of the subtangent to the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{9} = 1$ at a point where the x-coordinate is $x_1 = 2$:
Q3.The length of the subnormal at the endpoint of the major axis $(a, 0)$ evaluates to:
Q4.The ratio of the length of the subnormal $NG$ to the abscissa coordinate $x_1$ of any variable point on an ellipse is constant and equal to:
Q5.Find the length of the subtangent to the ellipse $x^2 + 4y^2 = 4$ at the parameter point parameterized at an eccentric angle $\theta = \frac{\pi}{3}$:
10
Module 10

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}

✎ Self-Check — 5 questions0 / 5
Q1.A light ray originates from the negative focus $S'(-4, 0)$ of the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{9} = 1$ and strikes the boundary curve at a point $P$ with an abscissa $x_1 = 3$. After reflecting off the surface, the ray travels along a line passing through:
Q2.The reflection property of the ellipse ensures that the normal line at any point $P$ on the curve always:
Q3.If a sound wave originates from one focus of a whispering gallery modeled by an elliptical room, the wave reflects off the wall and converges at:
Q4.The tangent line drawn at any point $P$ on an ellipse forms equal angles with:
Q5.A ray of light passing through the focus $S(3,0)$ of the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{16} = 1$ strikes the curve at the upper endpoint of the latus rectum. Find the length of the path traveled by the ray from $S$ to the curve and then to the other focus $S'$:
11
Module 11

Special Results and JEE Tricks

Special Results and JEE TricksTopic 1

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\begin{warningbox}[JEE Important Results — Ellipse]
  1. Focal Product Property: For any point $P(x_1, y_1)$ lying on the ellipse, the product of its focal distances satisfies: $SP \cdot S'P = a^2 - e^2 x_1^2$.
  2. The locus of the foot of the perpendicular dropped from either focus to any variable tangent line is the auxiliary circle $x^2 + y^2 = a^2$.
  3. The equation of the tangent lines drawn at the endpoints of the latus rectum can be simplified directly to the normal format: $\pm ex \pm \ds\frac{y}{a} = 1$.
  4. 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: $\ds\frac{\text{Area}_{\text{ellipse}}}{\text{Area}_{\text{circle}}} = \frac{b}{a}$.
\end{warningbox}
✎ Self-Check — 5 questions0 / 5
Q1.Find the locus of the foot of the perpendicular dropped from the focus $S(3,0)$ of the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{16} = 1$ to any variable tangent line:
Q2.A triangle inscribed inside the auxiliary circle $x^2 + y^2 = 16$ has an area of $12$ square units. Find the area of the corresponding mapped triangle inscribed inside the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{9} = 1$:
Q3.Find the value of the focal distance product $SP \cdot S'P$ for a point lying on the ellipse $\ds\frac{x^2}{25} + \frac{y^2}{16} = 1$ at the vertex position $(5, 0)$:
Q4.Find the equation of the tangent line to the ellipse $\ds\frac{x^2}{16} + \frac{y^2}{12} = 1$ at the upper endpoint of its latus rectum in the first quadrant (where $ae = 2$ and $\frac{b^2}{a} = 3$):
Q5.If the tangent line at any point $P$ on an ellipse meets the vertical directrix at $T$, the line segment $ST$ joining the focus to that intersection point always subtends what angle at the focus?
12
Module 12

Practice Problems (JEE Level)

Practice Problems (JEE Level)Topic 1

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  1. 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).
  2. 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 $b^2$.
  3. 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 $(2, 1)$.
  4. 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 $\ds\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
  5. Show that the sum of the squares of the lengths of any two conjugate semi-diameters of an ellipse is constant and equal to $a^2 + b^2$.
\subsubsection*{Practice Problems Solution Matrix (MCQ Wrappers)}
  1. 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?
    1. $a^2l^2 + b^2m^2 = 1$
    2. $a^2l^2 - b^2m^2 = 1$
    3. $a^2m^2 + b^2l^2 = 1$
    4. $l^2 + m^2 = a^2 + b^2$
  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 $e = 1/2$):
    1. $8$
    2. $4$
    3. $16$
    4. $2$
  3. Find the equation of the chord of $\ds\frac{x^2}{25} + \frac{y^2}{9} = 1$ bisected exactly at $(2, 1)$:
    1. $18x + 25y = 61$
    2. $9x + 25y = 43$
    3. $18x + 25y = 89$
    4. $2x + y = 5$
  4. 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:
    1. $13$
    2. $5$
    3. $25$
    4. $9$
  5. 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:
    1. $ab$
    2. $2ab$
    3. $\frac{1}{2}ab$
    4. $a^2 + b^2$
Answer Key: 1. (A) 2. (A) 3. (A) 4. (A) 5. (A)

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13
Module 13

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 ElementAlgebraic 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:$m_1 \cdot m_2 = -b^2/a^2$
\end{formulabox}

14
Module 14

Solved Examples

Worked ExamplesTopic 1

Worked Examples
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
Here $a = 3$ and $b = 2$. Evaluate the parameter coordinates: \[ x_1 = a\cos\left(\frac{\pi}{3}\right) = 3\left(\frac{1}{2}\right) = \frac{3}{2}, \quad y_1 = b\sin\left(\frac{\pi}{3}\right) = 2\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}. \]
  • 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: \[ 3x\sec\left(\frac{\pi}{3}\right) - 2y\csc\left(\frac{\pi}{3}\right) = a^2 - b^2 \implies 3x(2) - 2y\left(\frac{2}{\sqrt{3}}\right) = 9 - 4 \] \[ 6x - \frac{4}{\sqrt{3}}y = 5 \implies 6\sqrt{3}x - 4y - 5\sqrt{3} = 0. \]
Final Answer: Tangent: $2x + 3\sqrt{3}y - 12 = 0$; Normal: $6\sqrt{3}x - 4y - 5\sqrt{3} = 0$.
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Example 2 — Director Circle Locus Tracking

Find the locus of the point of intersection of two mutually perpendicular tangent lines drawn to the ellipse $9x^2 + 16y^2 = 144$.

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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 $x^2 + y^2 = 25$.

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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 $M(2, 1)$.

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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 $32x + 25y - 89 = 0$.