All points \(P\) satisfying \(PF_{1}+PF_{2}=k\), with fixed foci \(F_{1},F_{2}\) and constant \(k > F_{1}F_{2}\).
If the sum equalled the focal distance, what locus would form?
Follow the algebraic trail from definition to equation.
Place foci at \(F_1(-c,0)\) and \(F_2(c,0)\). Pick point \(P(x,y)\). Centre at origin, major axis on \(x\)-axis.
Compute \(PF_1=\sqrt{(x+c)^2+y^2}\) and \(PF_2=\sqrt{(x-c)^2+y^2}\). By definition, \(PF_1+PF_2=2a\).
Square twice and simplify to get \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \(b^2 = a^2 - c^2\).
Remember \(a \gt b\). If \(c=0\), the ellipse collapses into a circle.
Square each focus distance and add; constant \(2a\) fixes ellipse size.
Expand brackets, then isolate \(x\) and \(y\) terms; constants shift right.
Divide by \(a^2\) and set \(b\); standard ellipse form emerges.
Each algebra step mirrors geometry: squaring fixes focus distances, isolation separates variables, \(a\) and \(b\) reveal semi-axes.
Relation linking semi-major \(a\), semi-minor \(b\), and focal distance \(c\).
Compute \(c=\sqrt{a^{2}-b^{2}}\) when axes lengths are known.
Verify if three given lengths can form an ellipse by testing \(a^{2}=b^{2}+c^{2}\).
Supports proof that any point \(P\) satisfies \(PF_{1}+PF_{2}=2a\).
Eccentricity \(e\) quantifies ellipse flatness: \(0 < e < 1\). \(e\) near 0 ⇒ almost circle; \(e\) near 1 ⇒ very elongated.
Earth’s orbit has \(e \approx 0.017\) — practically circular.
The latus rectum is a special chord of an ellipse.
It passes through a focus and is perpendicular to the major axis.
Key facts in one glance
Set of all points whose distances to two foci add to \(2a\).
\(\displaystyle \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\; a>b>0\).
Semi-axes satisfy \(a^{2}=b^{2}+c^{2}\).
\(e=\frac{c}{a},\; 0<e<1\) gauges “ovalness”.
Through a focus, length \( \frac{2b^{2}}{a}\).