Ellipse — Foci, Eccentricity & Focal Distances

For $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ with $a > b$ (major axis along $x$):

• Foci: $(\pm c, 0)$ with $c^2 = a^2 - b^2$; distance between foci $= 2c$.
• Eccentricity: $e = \dfrac{c}{a} = \sqrt{1 - \dfrac{b^2}{a^2}}$ ($0 < e < 1$).
• Major axis is along the variable with the LARGER denominator. If $b > a$, swap roles: the major axis is along $y$, foci at $(0, \pm c)$.
• Parametric point: $(a\cos\theta,\ b\sin\theta)$ — eliminate $\theta$ with $\cos^2\theta+\sin^2\theta=1$ to recover the Cartesian equation, e.g. $(3\cos\theta,\,5\sin\theta)$ lies on $\tfrac{x^2}{9}+\tfrac{y^2}{25}=1$, major axis along $y$.
• A right angle at a focus — the perpendicularity test: two segments from a focus $S$ to points $P,Q$ are perpendicular iff $\vec{SP}\cdot\vec{SQ}=0$ (equivalently the slopes multiply to $-1$). Standard result: if the ends of the minor axis $(0,\pm b)$ subtend $90°$ at a focus $(ae,0)$, then $a^2e^2-b^2=0\Rightarrow b=c\Rightarrow a^2=2b^2$, so $e=\tfrac{1}{\sqrt2}$ — two separate PYQs ask for exactly this number. If the ends of the LATUS RECTUM subtend $90°$ at the centre, then $\tfrac{b^2}{a}=c$, giving $e^2+e-1=0$, $e=\tfrac{\sqrt5-1}{2}$.

For any point $P$ on the ellipse with foci $F_1, F_2$:

• This is the locus definition: 'sum of distances from two fixed points is constant'.
• The latus rectum has length $\dfrac{2b^2}{a}$, with endpoints at $\left(\pm c, \pm \tfrac{b^2}{a}\right)$.
• Each focal distance on its own: for $P(x_1,y_1)$ on the ellipse, $PF_1=a+ex_1$ and $PF_2=a-ex_1$ (focus $F_1$ at $(-c,0)$, $F_2$ at $(c,0)$) — from the focus–directrix definition, $PF=e\times$(distance to the directrix $x=\pm a/e$). They sum to $2a$ as they must, and a question that names ONE focus wants this, not the sum: on $\tfrac{x^2}{25}+\tfrac{y^2}{9}=1$ ($e=\tfrac45$) the point $(3,\tfrac{12}5)$ is $5+\tfrac{12}5=7.4$ from one focus and $2.6$ from the other.

Translate the given data into equations in $a, b, c$ (using $c = ae$, $c^2 = a^2 - b^2$, latus rectum $= 2b^2/a$):

• Vertices and foci given: read $a$ from the vertices, $c$ from the foci, then $b^2 = a^2 - c^2$.
• Two points given: substitute into $\tfrac{x^2}{A} + \tfrac{y^2}{B} = 1$ and solve the linear system in $\tfrac1A, \tfrac1B$.
• Position of a point $(x_1,y_1)$: inside if $\tfrac{x_1^2}{a^2}+\tfrac{y_1^2}{b^2} < 1$, on if $=1$, outside if $>1$.
• Area enclosed $= \pi a b$.