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`
`color(blue)"key Point :"` hyperbola in which a = b is called an equilateral hyperbola.
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.")`
`color(blue)"key Point :"` The standard equations of hyperbolas have transverse and conjugate axes as the coordinate axes and the centre at the origin. However, there are hyperbolas with any two perpendicular lines as transverse and conjugate axes, but the study of such cases will be dealt in higher classes.
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.`
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`
`color(blue)"key Point :"` hyperbola in which a = b is called an equilateral hyperbola.
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.")`
`color(blue)"key Point :"` The standard equations of hyperbolas have transverse and conjugate axes as the coordinate axes and the centre at the origin. However, there are hyperbolas with any two perpendicular lines as transverse and conjugate axes, but the study of such cases will be dealt in higher classes.
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.`