Fig 10.20 | Ellipse with foci \(F_1,F_2\); blue segments keep constant sum.
Definition 4: An ellipse is the locus of points whose distances to two foci add to a constant.
Thus, for every point \(P\), \(PF_1 + PF_2 = 2a\), a fixed length greater than \(F_1F_2\).
Fig 10.21 — Ellipse showing major and minor axes
Look at Fig 10.21 and match each label. After this, you should name the centre, vertices, major and minor axes with ease.
In Fig 10.22 and Fig 10.23, three numbers fully describe any ellipse drawn with centre O.
Know them, and you can link the semi-axes to the focus distance.
Place P on the major axis; ellipse definition gives the sum as the full length \(2a\).
Pick Q on the minor axis; each focal distance forms a right triangle, giving \(\sqrt{b^{2}+c^{2}}\).
Since the sum is constant for every point, equate results from P and Q.
Square both sides to relate semi-major, semi-minor, and focal lengths.
Comparing sums at symmetric points links the ellipse’s constant distance property directly to \(a\), \(b\), and \(c\).
Fig 10.24 • Ellipse with centre at O(0,0)
A centred ellipse has two orientations decided by its major axis.
Here \(a\) is the semi-major length and \(b\) the semi-minor, with \(a > b\).
Fig 10.25 – Point P on an ellipse with foci \(F_1,F_2\)
Fig 10.25 places \(P(x,y)\) on an ellipse whose foci are \(F_1(-c,0)\) and \(F_2(c,0)\).
By definition, the sum of distances satisfies \(PF_1 + PF_2 = 2a\).
The latus rectum is the chord that passes through a focus and is perpendicular to the major axis.
Knowing its length lets us quickly compare the “width” of different ellipses at their foci.
Example 9 — Find foci, vertices, eccentricity and latus rectum of \( \frac{x^{2}}{25}+\frac{y^{2}}{9}=1 \).
Compare with \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \) to get \(a=5,\,b=3\).
\(c^{2}=a^{2}-b^{2}=25-9=16 \Rightarrow c=4\).
Major axis lies on \(x\)-axis ⇒ foci at \((\pm4,0)\); vertices at \((\pm5,0)\).
\(e=\frac{c}{a}=\frac{4}{5}=0.8\); length of latus rectum \(=\frac{2b^{2}}{a}=\frac{18}{5}\).
For any ellipse centred at the origin, remember \(c^{2}=a^{2}-b^{2}\).
Use vertices & foci to model each ellipse.
Vertices at \((\pm13,0)\) give \(a=13\). Foci at \((\pm5,0)\) give \(c=5\). Then \(b=\sqrt{a^{2}-c^{2}}=\sqrt{169-25}=12\). Equation: \(\dfrac{x^{2}}{169}+\dfrac{y^{2}}{144}=1\).
Major axis length \(20\Rightarrow a=10\). Foci distance \(5\Rightarrow c=5\). Hence \(b^{2}=a^{2}-c^{2}=100-25=75\). Equation: \(\dfrac{x^{2}}{75}+\dfrac{y^{2}}{100}=1\).
Remember: \(b^{2}=a^{2}-c^{2}\) for every ellipse — your shortcut to quick equations.
Identify the standard form of an ellipse whose major axis lies on the y-axis.
The larger denominator shows the major axis. In Exercise 10.3 intro we learnt: if it is under \(y^{2}\), the axis is vertical.
Well done! The larger denominator 25 under \(y^{2}\) confirms a vertical major axis.
Remember: the ellipse stretches along the axis whose denominator is larger. Try again!
An ellipse is the set of points whose distances to two foci sum to a constant.
Semi-major \(a\), semi-minor \(b\), focal length \(c\) satisfy \(c^{2}=a^{2}-b^{2}\).
Standard forms: \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \) and \( \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1 \).
Eccentricity \(e=\frac{c}{a}\); length of latus rectum \( \frac{2b^{2}}{a} \).
Use solved examples, then attempt the practice set to consolidate learning.
Review the summary, finish the practice set, and bring questions to the next class.
Thank You!
We hope you found this lesson informative and engaging.