`\color{fuchsia}(ul "★ Hyperbola ")`

A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant.

`\color{blue} ✍️` The term “difference” that is used in the definition means the distance to the further point minus the distance to the closer point. The two fixed points are called `color(purple)("the foci of the hyperbola.")`

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

`\color{blue} ✍️` The line through the foci is called `color(blue)("the transverse axis")` and the line through the centre and perpendicular to the transverse axis is called `color(blue)("the conjugate axis.")`

`\color{blue} ✍️` The points at which the hyperbola intersects the transverse axis are called `color(blue)("the vertices of the hyperbola")` (Fig 11.29).

We denote the distance between the two foci by `2c,` the distance between two vertices (the length of the transverse axis) by `2a` and we define the quantity `b` as

`b = sqrt(c^2 - a^2)`

Also `2b` is the length of the conjugate axis (Fig 11.30).

`ul"To find the constant" \ \P_1F_2 – P_1F_1 :`

By taking the point `P` at `A` and `B` in the Fig 11.30, we have

`BF_1 – BF_2 = AF_2 – AF_1` (by the definition of the hyperbola)

`BA +AF_1– BF_2 = AB + BF_2– AF_1`

i.e., `AF_1 = BF_2`

So that, `color(red)(BF_1 – BF_2 )= BA + AF_1– BF_2 = BA = color(red)(2a)`

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

Just like an ellipse, the ratio `color(red)(e = c/a)` is called `color(blue)("the eccentricity of the hyperbola.")`

Since `c ≥ a,` the eccentricity is never less than one.

`\color{blue} ✍️` In terms of the eccentricity, the foci are at a distance of `color(red)(ae)` from the centre.

The equation of a hyperbola is simplest if the centre of the hyperbola is at the origin and the foci are on the `x`-axis or `y`-axis. The two such possible orientations are shown in Fig `11.31.`

`\color{blue} ✍️` `ul"We will derive the equation for the hyperbola shown in Fig 11.31(a)"` with foci on the x-axis.

Let `F_1` and `F_2` be the foci and O be the mid-point of the line segment `F_1F_2.`

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

The line through `O` perpendicular to the x-axis be the y-axis.

Let the coordinates of `color(blue)(F_1)` be `color(blue)((– c,0))` and `color(blue)(F_2)` be `color(blue)((c,0))` (Fig 11.32).

Let `P(x, y)` be any point on the hyperbola such that the difference of the distances from `P` to the farther point minus the closer point 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`

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

Squaring both side, we get

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

and on simplifying, we get

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

On squaring again and further simplifying, we get

`(x^2)/(a^2) - (y^2)/(c^2-a^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 hyperbola satisfies")`

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

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

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

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

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

Similarly, `PF_2 = a - a/c x`

In hyperbola `c > a;` and since P is to the right of the line `x = a, x > c/a x > a` Therefore,

`a-c/ax` becomes negative Thus, `PF_2 = c/a x -a`

Therefore `PF_1 – PF_2 = a + c/ax - (cx)/a + a= 2a`

Also, note that if P is to the left of the line `x = – a,` then

`PF_A=-(A+c/ax) PF_2 = a- c/ax`

In that case `P F_2 – PF_1 = 2a`.

So, any point that satisfies `(x^2)/(a^2) - (y^2)/(b^2) = 1` lies on the hyperbola.

Thus, we proved that the equation of hyperbola with origin (0,0) and transverse axis along `x`-axis is `(x^2)/(a^2) - (y^2)/(b^2) =1`



Similarly, we can derive the equation of the hyperbola in `color(blue)(Fig 11.31 (b))` as `color(red)((y^2)/(a^2) - (x^2)/(b^2) = 1)`

These `color(blue)(ul"two equations are known as the standard equations of hyperbolas.")`



From the standard equations of hyperbolas (Fig11.29),

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

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

`color{maroon}{2.}` The foci are always on the transverse axis. It is the positive term whose denominator gives the transverse axis.

`color(red)("For example")`, `(x^2)/9 - (y^2)/16 = 1` has transverse axis along `x`-axis of length `6`,

while `(y^2)/25 - (x^2)/16 = 1` has transverse axis along y-axis of length `10.`

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

Latus rectum of hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose end points lie on the hyperbola.

As in ellipse, it is easy to show that `color(red)(ul"the length of the latus rectum in hyperbola is" \ \(2b^2)/a)`