Please Wait... While Loading Full Video#### Class 11 Chapter 11 - Conic Sections

### Hyperbola

`star` Hyperbola

`star` Eccentricity

`star` Standard equation of Hyperbola

`star` Latus rectum

`star` Eccentricity

`star` Standard equation of Hyperbola

`star` Latus rectum

`\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)`

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.

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{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.`

`\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)`

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)`

Q 3185091867

Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas:

`(i) x^2/9 - (y^2)/16 = 1`

`(ii) y^2 - 16x^2= 16`

`(i) x^2/9 - (y^2)/16 = 1`

`(ii) y^2 - 16x^2= 16`

(i) Comparing the equation `x^2/9 - (y^2)/16 = 1` with the standard equation

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

Here, `a = 3, b = 4` and `c = sqrt(a^2+b^2) = sqrt(9+16)=5`

Therefore, the coordinates of the foci are `(± 5, 0)` and that of vertices are `(± 3, 0).`Also,

The eccentricity `e = c/a = 5/3` The latus rectum `= (2b^2)/a = 32/3`

(ii) Dividing the equation by 16 on both sides, we have `y^2/16 - x^2/1 = 1`

Comparing the equation with the standard equation `y^2/a^2 - x^2/b^2 = 1` we find that

`a = 4, b = 1` and `c=sqrt(a^2+b^2) = sqrt(16+1) = sqrt(17)`

Therefore, the coordinates of the foci are `(0, ± sqrt17)` and that of the vertices are `(0, ± 4).` Also,

The eccentricity `e = c/a sqrt17/4` The rectum `= (2b^2)/a^2 = 1/2`

Q 3146101073

Find the equation of the hyperbola with foci `(0, ± 3)` and vertices `(0 ± sqrt11/2)`

Since the foci is on y-axis, the equation of the hyperbola is of the form

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

Since vertices are `(0,± sqrt11/2 ) a = sqrt11/2`

Also, since foci are `(0, ± 3); c = 3 and b2 = c2 – a2 = 25/4.`

Therefore, the equation of the hyperbola is

`y^2/((11/4)) - x^2/((25/4)) = 1, i.e 100y^2 - 44x^2 = 275`

Q 3166101075

Find the equation of the hyperbola where foci are `(0, ±12)` and the length of the latus rectum is `36.`

Since foci are `(0, ± 12),` it follows that `c = 12.`

Length of the latus rectum `= (2b^2)/a = 36` or `b^2 = 18a`

Therefore `c^2 = a^2 + b^2;`gives

`144 = a^2 + 18a`

i.e., `a^2 + 18a – 144 = 0,`

So `a = – 24, 6.`

Since a cannot be negative, we take `a = 6` and so `b^2 = 108.`

Therefore, the equation of the required hyperbola `y^2/36 - x^2/108 = 1i.e 3y^2-x^2 = 108`