#### notes

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 in following fig.

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 `F_1` be (– c,0) and `F_2` be (c,0) in following fig.

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`

i.e., `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`)

Hence any point on the hyperbola satisfies `x^2/a^2 -y^2/b^2=1`

From the standard equations of hyperbolas, we observ that:

1. 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.

2. The foci are always on the transverse axis. It is the positive term whose denominator gives the transverse axis.