`\color{blue} ✍️` An `color{green}ul"ellipse"` is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is a constant. The two fixed points are called the foci (plural of ‘focus’) of the ellipse (Fig11.20).

`\color{blue} ✍️` The mid point of the line segment joining the foci is called `color(blue)("the centre of the ellipse.")`

`\color{blue} ✍️` The line segment through the foci of the ellipse is called `color(blue)("the major axis")` and the line segment through the centre and perpendicular to the major axis is called `color(blue)("the minor axis.")`

`\color{blue} ✍️` The end points of the major axis are called the vertices of the ellipse(Fig 11.21).

`\color{blue} ✍️` We denote the length of the major axis by `2a,` the length of the minor axis by `2b` and the distance between the foci by `2c`. Thus, the length of the semi major axis is `a` and semi-minor axis is `b`. (Fig11.22).

`\color{blue} ✍️` Take a point `P` at one end of the major axis.

`\color{blue} ✍️` Sum of the distances of the point P to the foci is `color(green)(F_1 P + F_2P = F_1O + OP + F_2P)`

(Since, `F_1P = F_1O + OP`)

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \= c + a + a – c = 2a`

`\color{green} ✍️` Take a point `Q` at one end of the minor axis. Sum of the distances from the point `Q` to the foci is

`F_1 Q +F_2Q = sqrt(b^2+c^2) + sqrt(b^2 +c^2) = 2sqrt(b^2+c^2)`

`\color{green} ✍️` Since both P and Q lies on the ellipse. By the definition of ellipse, we have

`sqrt2(b^2+c^2) = 2a.` i.e ` \ \ \ \ \ \ \ \ color(red)(a= sqrt(b^2+c^2))`

`a^2=b^2+c^2,` i.e ` \ \ \ color(red)("the distance of the focus from the centre of the ellipse"\ \ c =sqrt(a^2-b^2))`

In the equation `c^2 = a^2 – b^2` obtained above, if we keep a fixed and vary c from 0 to a, the resulting ellipses will vary in shape.

`color(red)"Case (i)"` When `color(blue)(c = 0)`, both foci merge together with the centre of the ellipse and `a^2 = b^2,` i.e., `color(red)(a = b)`, and so the ellipse becomes circle (Fig11.24).

`color(red)"Case (ii)"` When `color(blue)(c = a)`, then `color(blue)(b = 0.)` The ellipse reduces to the line segment `F_1F_2` joining the two foci (Fig11.25).

`\color{fuchsia}(ul "★ Eccentricity")`

The eccentricity of an ellipse is the ratio of the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse (`color(blue)"eccentricity is denoted by e")` i.e `color(red)(e = c/a)`

Then since the focus is at a distance of `c` from the centre, in terms of the eccentricity the focus is at a distance of `color(blue)(ae)` from the centre.

`\color{blue} ✍️` The equation of an ellipse is simplest if the centre of the ellipse is at the origin and the foci are on the x-axis or y-axis.

`\color{blue} ✍️` The two such possible orientations are shown in Fig 11.26.

`\color{blue} ✍️` We will derive the equation for the ellipse shown above in Fig 11.26 (a) with foci on the `x`-axis.

`\color{blue} ✍️` Let `F_1` and `F_2` be the foci and `O` be the midpoint of the line segment `F_1F_2`.

Let `O` be the origin and the line from `O` through `F_2` be the positive x-axis and that through `F_1` as the negative `x`-axis.

Let, the line through `O` perpendicular to the x-axis be the y-axis. Let the coordinates of `F_1` be `(– c, 0)` and `F_2` be `(c, 0)` (Fig 11.27).

Let `P(x, y)` be any point on the ellipse such that the sum of the distances from `P` to the two foci be `2a` so given

`PF_1 + PF_2 = 2a`

Using the distance formula, we have

`sqrt((x+c)^2 + y^2) + sqrt((x-c)^2+y^2)=2a`

i.e `sqrt((x+c)^2 +y^2) = 2a - sqrt((x-c)^2) +y^2`

Squaring both sides, we get

`(x + c)^2 + y^2 = 4a^2 – 4a sqrt((x-c)^2 + y^2)+(x -c)^2 +y^2`

which on simplification gives

`sqrt((x-c)^2 +y^2) = a-c/a x`

Squaring again and simplifying, we get

`(x^2)/(a^2) + (y^2)/(a^2-c^2) = 1`

i.e `(x^2)/(a^2) + (y^2)/(b^2) = 1` ` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ "Since" (c^2 = a^2 – b^2)`

`color(blue)("Hence any point on the ellipse satisfies")`

`color(blue)((x^2)/(a^2) + (y^2)/(b^2) = 1)`.......................................(2)

`\color{blue} ✍️` `color(red)("Conversely,")` let `P (x, y)` satisfy the equation `(2)` with `0 < c < a`. Then

`y^2 = b^2 (1-(x^2)/(a^2))`

Therefore, `PF_1 = sqrt((x+c)^2 +y^2)`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =sqrt((x+c)^2+b^2(a^2-x^2)/(a^2))`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =sqrt((x+c)^2+(a^2-c^2) (a^2-x^2)/(a^2))` ` \ \ \ \ \ \ \ \ \ \("since" b^2 = a^2 – c^2)`

`\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sqrt((a+(cx)/a)^2) = a+c/a x`

Similarly `PF_2 = a- c/ax`

Hence `PF_1 + PF_2 = a+c/a x + a-c/a x = 2a`........................................(3)

So, any point that satisfies `(x^2)/(a^2) + (y^2)/(b^2) = 1` satisfies the geometric condition and so
P `(x, y)` lies on the ellipse.

Hence from (2) and (3), we proved that the equation of an ellipse with centre of the origin and major axis along the x-axis is

`(x^2)/(a^2) + (y^2)/(b^2) = 1`

Similarly, we can derive the equation of the ellipse in Fig 11.26 (b) as `color(red)((x^2)/(b^2) + (y^2)/(a^2) =1)`

`\color{orange}{"These two equations are known as standard equations of the ellipses."}`



From the standard equations of the ellipses (Fig11.26),

`color{maroon}{ul"we have the following observations:"}`

`color{maroon}{1.}` `\color{orange}{"Ellipse is symmetric with respect to both the coordinate axes"}` since if `(x, y)` is a point on the ellipse, then `(– x, y), (x, –y)` and `(– x, –y)` are also points on the ellipse.

`color{maroon}{2.}` The foci always lie on the major axis. The major axis can be determined by finding the intercepts on the axes of symmetry. That is, major axis is along the x-axis if the coefficient of `x^2` has the larger denominator and it is along the y-axis if the coefficient of `y^2` has the larger denominator.

`\color{fuchsia}(ul "★ Latus rectum ")`

Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse (Fig 11.28).

`ul"To find the length of the latus rectum of the ellipse"` `(x^2)/(a^2) + (y^2)/(b^2) = 1`

Let the length of `AF_2` be `l.`

Then the coordinates of `A` are `(c, l )` ,i.e., ` \ \ \ \(ae, l )`

Since A lies on the ellipse `(x^2)/(a^2) +(y^2)/(b^2) = 1` have

` \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ((ae)^2)/(a^2) - (l^2)/(b^2) = 1`

`=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ l^2 = b^2 (1-e^2)`

But `\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ e^2 = (c^2)/(a^2) = (a^2-b^2)/(a^2) = 1- (b^2)/(a^2)`

Therefore `l^2 = (b^4)/(a2),` i.e ` \ \ \ \ \ \ \ color(red)(l = (b^2)/a)`

Since the ellipse is symmetric with respect to `y`-axis (of course, it is symmetric w.r.t.
both the coordinate axes), `AF_2 = F_2B` and

so `color(red)("length of the latus rectum is" \ \ \ \ (2b^2)/a)`