What is the definition of an ellipse in Class 11?

An ellipse is the locus of points in a plane whose sum of distances from two fixed points (called foci) is constant. For the standard form x^2/a^2 + y^2/b^2 = 1, this sum equals 2 a.

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Ellipse – Definition, Standard Equation and Properties

Q: What is the length of the latus rectum of an ellipse?

Length of latus rectum = 2 b squared / a, where 2 a is the major axis and 2 b the minor axis.

Q: How are planetary orbits related to ellipses?

Kepler's first law states planets move in elliptical orbits with the Sun at one focus. Earth's orbit has a small eccentricity of about 0.0167.

Q: Can an ellipse be a circle?

Yes, when a = b the ellipse equation becomes x squared + y squared = a squared, which is a circle. The circle is an ellipse with eccentricity 0.

🎓 Class 11 Mathematics CBSE Theory Ch 10 — Conic Sections ⏱ ~30 min

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10.5 Ellipse

Definition

An ellipseⁱ is the locus of all points in a plane the sum of whose distances from two fixed points (the foci) is a constant (greater than the distance between the foci).

10.5.1 Terminology

Let \(F_1, F_2\) be the foci and let \(2c = F_1F_2\). For a point \(P\) on the ellipse, \(PF_1+PF_2=2a\) with \(2a>2c\).

Centre \(O\): midpoint of \(F_1F_2\).
Major axis — line segment through the foci; length \(2a\). Vertices are the endpoints of the major axis.
Minor axis — perpendicular to the major axis at \(O\); length \(2b\). The semi-minor-axis \(b\) satisfies \(b^2=a^2-c^2\).
Eccentricity \(e=\dfrac{c}{a},\ 0

10.5.2 Derivation of the standard equation

Place the centre at origin and the foci on the \(x\)-axis at \(F_1(-c,0),F_2(c,0)\). Let \(P(x,y)\) be on the ellipse.

\(PF_1+PF_2=2a\Rightarrow\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2a.\)

Isolate one radical, square, simplify, isolate the remaining radical and square again; using \(b^2=a^2-c^2\) one arrives at:

Standard form (foci on x-axis)

\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a>b>0,\quad b^2=a^2-c^2.\]

If the foci lie on the \(y\)-axis, the standard form is \(\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1\) with \(a>b>0\).

Fig 10.6 — Ellipse with foci \(F_1,F_2\) on the \(x\)-axis; \(PF_1+PF_2=2a\).

10.5.3 Eccentricity and latus rectum

The eccentricityⁱ \(e=c/a\) measures the "flatness" of the ellipse. When \(e\) is close to 0 the ellipse is nearly circular; as \(e\to 1\) it becomes increasingly elongated.

The latus rectum (chord through either focus perpendicular to the major axis): at \(x=\pm c\), \(\dfrac{c^2}{a^2}+\dfrac{y^2}{b^2}=1\Rightarrow y=\pm\dfrac{b^2}{a}\). Hence length \(=\dfrac{2b^2}{a}\).

Example 10

Find the coordinates of the foci, the vertices, the length of the major and minor axes, eccentricity and length of the latus rectum of \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\).
Here \(a^2=25,b^2=9\), so \(a=5,b=3\). \(c=\sqrt{a^2-b^2}=\sqrt{16}=4\). Foci \((\pm 4,0)\); vertices \((\pm 5,0)\); major axis \(=10\); minor axis \(=6\); \(e=\tfrac{4}{5}\); LR \(=\tfrac{2\cdot 9}{5}=\tfrac{18}{5}\).

Example 11

Find the standard equation of the ellipse with foci \((\pm 3,0)\) and vertices \((\pm 5,0)\).
\(a=5,c=3\Rightarrow b^2=25-9=16\). Equation: \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1.\)

Example 12

Find the equation of the ellipse with foci on the \(y\)-axis, centre at origin, passing through \((3,2)\) with eccentricity \(1/2\).
Form: \(\tfrac{x^2}{b^2}+\tfrac{y^2}{a^2}=1\). \(e=c/a=\tfrac12\Rightarrow c=a/2\Rightarrow b^2=a^2-c^2=\tfrac{3a^2}{4}\). At \((3,2)\): \(\tfrac{9}{3a^2/4}+\tfrac{4}{a^2}=1\Rightarrow \tfrac{12}{a^2}+\tfrac{4}{a^2}=1\Rightarrow a^2=16,b^2=12\). Equation: \(\dfrac{x^2}{12}+\dfrac{y^2}{16}=1.\)

Example 13

Find the eccentricity of the ellipse whose latus rectum equals half its major axis.
\(\tfrac{2b^2}{a}=\tfrac{2a}{2}=a\Rightarrow 2b^2=a^2\Rightarrow b^2=a^2/2\). \(c^2=a^2-b^2=a^2/2\Rightarrow e=c/a=\tfrac{1}{\sqrt 2}.\)

Activity 10.3 — The two-pin-and-string ellipse

Predict: If you push two pins 6 cm apart and loop a string of total length 10 cm over them, then trace taut with a pencil, what curve emerges?

Fix pins at \(F_1,F_2\) with \(F_1F_2=6\) cm — so \(2c=6,\ c=3\).
String of length \(2a=10\Rightarrow a=5\). Trace the taut loop.
Compute \(b=\sqrt{a^2-c^2}=\sqrt{16}=4\). The shortest diameter of the curve should be \(2b=8\) cm — measure and verify.
Compute \(e=c/a=3/5=0.6\). Note: the larger \(c\) (wider-apart pins), the more elongated the curve.

Insight: The locus is an ellipse: every traced point satisfies \(PF_1+PF_2=\) loop length \(=2a\). The physical construction captures the defining property exactly.

In-text Exercises on the Ellipse

Q1. Find all relevant quantities for \(\dfrac{x^2}{36}+\dfrac{y^2}{16}=1\).

\(a=6,b=4,c=\sqrt{20}=2\sqrt5\). Foci \((\pm 2\sqrt5,0)\); vertices \((\pm 6,0)\); major \(=12\); minor \(=8\); \(e=\tfrac{2\sqrt5}{6}=\tfrac{\sqrt5}{3}\); LR \(=\tfrac{32}{6}=\tfrac{16}{3}.\)

Q2. Find the equation of the ellipse with vertices \((0,\pm 13)\) and foci \((0,\pm 5)\).

Foci on \(y\)-axis, \(a=13,c=5\Rightarrow b^2=144\). Equation: \(\dfrac{x^2}{144}+\dfrac{y^2}{169}=1.\)

Q3. Find the equation of the ellipse centered at origin with major axis \(2a=20\) along \(x\)-axis and passing through \((3,-2)\).

\(a=10\). At \((3,-2)\): \(\tfrac{9}{100}+\tfrac{4}{b^2}=1\Rightarrow \tfrac{4}{b^2}=\tfrac{91}{100}\Rightarrow b^2=\tfrac{400}{91}\). Equation: \(\dfrac{x^2}{100}+\dfrac{91y^2}{400}=1.\)

Q4. Find the eccentricity of \(4x^2+9y^2=36\).

Divide by 36: \(\tfrac{x^2}{9}+\tfrac{y^2}{4}=1\Rightarrow a=3,b=2,c=\sqrt 5\Rightarrow e=\dfrac{\sqrt5}{3}.\)

Competency-Based Questions — Planetary Orbits

A satellite orbits Earth along an ellipse with Earth at one focus. The satellite's closest approach (perigee) is \(6800\) km from Earth's centre and its farthest point (apogee) is \(42200\) km. Assume Earth's centre is at one focus of the ellipse.

Q1. Compute the semi-major axis \(a\) and the distance \(c\) from centre to focus.

L3 Apply

Perigee + apogee = \(2a\Rightarrow 2a=49000\Rightarrow a=24500\) km. \(c=a-\text{perigee}=24500-6800=17700\) km.

Q2. Determine the semi-minor axis \(b\) and the eccentricity \(e\) of the orbit.

L4 Analyse

\(b=\sqrt{a^2-c^2}=\sqrt{24500^2-17700^2}=\sqrt{600{,}250{,}000-313{,}290{,}000}=\sqrt{286{,}960{,}000}\approx 16{,}940\) km. \(e=c/a=17700/24500\approx 0.723.\)

Q3. A second satellite is circular with radius \(a\). Evaluate which has greater total path length (perimeter) and justify qualitatively.

L5 Evaluate

For fixed semi-major axis \(a\), the circle has perimeter \(2\pi a\). Any ellipse with the same \(a\) but \(bcircle has the greater perimeter.

Q4. Design an elliptical orbit with eccentricity 0.5 and perigee 7000 km. State the equation (with Earth's centre at the +x focus and ellipse centred at origin).

L6 Create

Perigee \(=a-c=a(1-e)=0.5a=7000\Rightarrow a=14000\) km. \(c=7000\); \(b^2=a^2-c^2=196000000-49000000=147000000\Rightarrow b=\sqrt{147000000}\approx 12124\) km. Equation: \(\dfrac{x^2}{1.96\times10^8}+\dfrac{y^2}{1.47\times10^8}=1.\)

Assertion–Reason Questions

Assertion (A): For any point on an ellipse, the sum of the distances from the two foci is constant and equal to the length of the major axis.
Reason (R): The defining property of an ellipse is \(PF_1+PF_2=2a\).

A) Both true; R explains A

B) Both true; R doesn't explain

C) A true, R false

D) A false, R true

Answer: A. Major axis length \(=2a\); the defining sum equals exactly this.

Assertion (A): The eccentricity of the ellipse \(\tfrac{x^2}{16}+\tfrac{y^2}{25}=1\) is \(\tfrac{3}{5}\).
Reason (R): For an ellipse with foci on the \(y\)-axis, \(a^2=25,b^2=16\Rightarrow c^2=9\Rightarrow e=c/a=3/5\).

A) Both true; R explains A

B) Both true; R doesn't explain

C) A true, R false

D) A false, R true

Answer: A. The calculation in R is correct and supports A.

Assertion (A): As \(e\to 0\), the ellipse approaches a circle.
Reason (R): When \(c=0\), the two foci coincide and \(b=a\).

A) Both true; R explains A

B) Both true; R doesn't explain

C) A true, R false

D) A false, R true

Answer: A. \(e\to0\iff c\to0\iff F_1\!=\!F_2\iff b\to a\). The ellipse collapses to a circle.

Frequently Asked Questions

What is the standard equation of an ellipse?

The standard equation is x^2/a^2 + y^2/b^2 = 1. If a > b, the major axis is along the x-axis with length 2 a, and the minor axis along the y-axis with length 2 b.

Where are the foci of an ellipse located?

For the ellipse x^2/a^2 + y^2/b^2 = 1 (a > b), the foci are at (plus/minus c, 0) on the major axis, where c squared = a squared - b squared.

What is the eccentricity of an ellipse?

Eccentricity e = c/a, where c squared = a squared - b squared. For every ellipse 0 <= e < 1. A smaller e means a more circular ellipse.

What is the length of the latus rectum of an ellipse?

Length of latus rectum = 2 b squared / a, where 2 a is the major axis and 2 b the minor axis.

How are planetary orbits related to ellipses?

Kepler's first law states planets move in elliptical orbits with the Sun at one focus. Earth's orbit has a small eccentricity of about 0.0167.

Can an ellipse be a circle?

Yes, when a = b the ellipse equation becomes x squared + y squared = a squared, which is a circle. The circle is an ellipse with eccentricity 0.

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