Ellipse: An ellipse in the set of all points in a plane such that the sum of the distances from two fixed points in the plane is a constant.

$P F_1+P F_2=\text { constant. }$

Special case:

$F_1=F_2$

$P F_1+P F_2=2 r$

$P_1 F_1+P_1 F_2=\text { constant }$

The two fixed point are called the foci (plural of focus) of the ellipse.

centre: mid point on the line segment joining the foci.

Major axis: line segment joining two points of the ellipse passing through the foci

Minor axis: the line segment passing through the centre and perpendicular to the major axis.

Vertices: the end points of the mage axis.

$A B$ : major axis

$C D$ : minor axis

$F_1 F_2$ : foci

O : centre

Consider an ellipse whose foci lie on the $x$-axis and the centre is at the origin

Relationship between $a, b$ and $c$

$B F_1+B F_2 =(B O+O F_1)+(B O-O F_2)$

$=(a+c)+(a-c)$

$=2 a$

$C F_1 =\sqrt{b^2+c^2}$

$C F_2 =\sqrt{b^2+c^2}$

$\therefore C F_1 +C F_2=2 \sqrt{b^2+c^2}$

Since, $B F_1+B F_2=C F_1+C F_2$, we get

$2 a=2 \sqrt{b^2+c^2}$

$\Rightarrow b^2+c^2=a^2$

$\Rightarrow c^2=a^2-b^2$

$\Rightarrow c=\sqrt{a^2-b^2}$

$e=\frac{c}{a} :$

$\text { the ratio of the distance }$
$\text { between the foci and the }$
$\text { distance between the vertices. }$

$F_1 F_2=2 c$

$A B=2 a$

$\frac{F_1 F_2}{A B}=\frac{2 c}{2 a}=\frac{c}{a}$

So, $c=a e$

Note that since $\quad c

let’s take an ellipse with centre at the origin and foci on the $x$-axis.

$P F_1+P F_2=\text { constant. }$

$B F_1+B F_2=2 a$

So, $P F_1+P F_2=2 a$

$\Leftrightarrow \sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2 a$

$\Leftrightarrow \quad(x+c)^2+y^2={(2 a-\sqrt{(x-c^2+y^2}})^2$

$\Leftrightarrow \quad(x+c)^2+y^2$ $=4 a^2+(x-c)^2+y^2$ $-4 a \sqrt{(x-c)^2+y^2}$

$\Leftrightarrow 4 a \sqrt{(x-c)^2+y^2}$ $=4 a^2+(x-c)^2-(x+c)^2={-4 x c}$

$\Leftrightarrow 4 a \sqrt{(x-c)^2+y^2}=4(a^2-c x)$

$\Leftrightarrow a^2[(x-c)^2+y^2]=(a^2-c x)^2$ $=a^4-2 c a^2 x +c^2x^2$

$\Leftrightarrow a^2 x^2+a^2 c^2-2 a^2 c x+a^2 y^2$ $=a^4-2 a^2 cx+c^2 x^2$

$\Leftrightarrow (a^2-c^2)= x^2+a^2 y^2$ $=a^4-a^2 c^2=a^2 (a^2-c^2)$

$\Leftrightarrow b^2 x^2+a^2 y^2=a^2 b^2$

$\Leftrightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

[Symmetric about both x and y axis]

Equation of an ellipse whose vertices are at $(-a, 0)$and $(a, 0)$, centre at $(0,0)$.

length of major axis $=2 a$

length of minor axis $=2b$.

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

If $a=b$, then we get a circle.

$\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$

$\Leftrightarrow \quad x^2+y^2=a^2$

circle centred at $(0,0)$ and radius $a$.

Line segment with endpoints on the ellipse passing through a focus and perpendicular to the major axis.

Let $P F=Q F=l$.

He need to find $2 l$.

$P \equiv(c, l)=(a e, l)$

Since $P(a e, l)$ lies on the ellipse

$\frac{x^2}{a^2}+\frac{l^2}{b^2}=1$,

$\frac{a^2 e^2}{a^2}+\frac{l^2}{b^2}=1$

$\Rightarrow \frac{l^2}{b^2}=1-e^2=1-\frac{c^2}{a^2}$

$\Rightarrow \frac{l^2}{b^2}=\frac{a^2-c^2}{a^2}=\frac{b^2}{a^2}$

$\Rightarrow l^2=\frac{b^4}{a^2}$

$\Rightarrow l=\frac{b^2}{a}$

$$\Rightarrow 2 l=\frac{2 b^2}{a}=\text { length of latus rectum }$$

Find the foci, vertices, eccentricity and length of latus rectum for the ellipse

$16 x^2+y^2=16$.

Solution: $\quad \frac{x^2}{1^2}+\frac{y^2}{4^2}=1$

$\Rightarrow a=1, b=4$.

Here $a

$F_1 \equiv(0,-c)$

$F_2 \equiv(0, c)$

$e=\frac{c}{b}$

$c^2 =b^2-a^2$

$=4^2-1^2=15$

So,

$c=\sqrt{15}$

$e=\frac{c}{b}=\frac{\sqrt{15}}{4}$

Foci : $(0, \pm \sqrt{15})$

length of latus rectum

$\frac{2 a^2}{b}=\frac{2}{4}=\frac{1}{2}$

$16 x^2+(\sqrt{15})^2=16$

$16 x^2=1 \Rightarrow x=\frac{1}{4}$

$P Q=2 x=\frac{1}{2}$

Find the equation of the ellipse whose centre is at $(0,0)$, major axis in the $y$-axis, and passes through $(3,2)$ and $(1,6)$

Solution:

$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ $\rightarrow \text { ellipse centre at} (0,0)$ and major axis on the $y$ axis

since it passes through points $(3,2)$ and $(1,6)$, we get

$\frac{9}{b^2}+\frac{4}{a^2}=1 \quad \text {-(i) }$

$\frac{1}{b^2}+\frac{36}{a^2}=1 \quad \text {-(ii) }$

$9 \times(i)- (ii)$

$\Rightarrow \frac{80}{b^2}=8$$\quad$ $\Rightarrow b^2=10$

$\therefore \text { (i) } \Rightarrow \frac{9}{10}+\frac{4}{a^2}=1$

$\Rightarrow \frac{4}{a^2}=\frac{1}{10}$$\quad$ $\Rightarrow a^2=40$

$\therefore$ the equation is $\frac{x^2}{10}+\frac{y^2}{40}=1$

Thank you