**BASIC CONCEPT:**

A hyperbola is the locus of a point which moves such that, ratio of its distance from a fixed point ( **focus **) and its distance from a fixed straight line ( **directrix **) is a constant ( **eccentricity **).

**This constant (eccentricity) is greater than unity.**

## Standard Equation and Basic Definitions:

Let S be the focus and ZM the directrix of a hyperbola.

Since e > 1, we can divide SZ internally and externally in the ratio e : 1, let the points of division be A and A’ as in the figure.

Let AA’ = 2a and C is the mid point of AA’

Then, SA = e. AZ

and SA’ = e. ZA’

⇒ SA + SA’ = e(AZ + ZA’)

= 2ae

i.e., 2SC = 2ae or SC= ae.

Similarly by subtraction,

SA’ − SA = e(ZA’ − ZA) = 2e(ZC)

⇒ 2a = 2e(ZC)

⇒ ZC = a/e

Now, take C as the origin , CS as the x-axis, and the perpendicular line CY as the y axis.

Then, S is the point (ae , 0) and ZM the line x = a/e.

### Let P(x , y) be any point on the hyperbola

Then the condition PS^{2} = e^{2}. (distance of P from ZM)^{2} gives

(x − ae)^{2} + y^{2} = e^{2} (x − a/e)^{2}

x^{2}(1 – e)^{2} + y^{2} = a^{2}(1 – e^{2})

$ \displaystyle \frac{x^2}{a^2} – \frac{y^2}{a^2(e^2 -1)} = 1 $ …(i)

Since e > 1, e^{2} – 1 is positive.

Let a^{2}(e^{2} − 1) = b^{2} .

Then the equation ( i ) becomes

$ \displaystyle \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 $

∎ The eccentricity e of the hyperbola $ \displaystyle \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 $ is given by the relation

$ \displaystyle e^2 = (1+ \frac{b^2}{a^2}) $

∎ Since the curve is symmetrical about the y – axis, it is clear that there exists another focus S’ at (−ae, 0) and a corresponding directirx Z’M’ with the equation x= −a/e , such that the same hyperbola is described if a point moves so that its distance from S’ is e times its distance from Z’M’.

∎ The points A and A’ where the straight line joining the two foci cuts the hyperbola are called the vertices of the hyperbola.

∎ The straight line joining the vertices is called the transverse axis of the hyperbola, its length AA’ is 2a

∎ The middle point C of AA’ possesses the property that it bisects every chord of the hyperbola passing through it. It can be proved by taking P(x_{1} , y_{1}) as any point on hyperbola.

If (x_{1} , y_{1}) lies on the hyperbola then so does P'(−x_{1} , −y_{1}) because hyperbola is symmetric about x and y axes. Therefore PP’ is a chord whose middle point is (0, 0), i.e. the origin O. On account of this property the middle point of the straight line joining the vertices of the hyperbola is called the center of the hyperbola.

∎ The straight line through the center of a hyperbola which is perpendicular to the transverse axis does not meet the hyperbola in real points. If B and B’ be the points on this line such that BC = CB’ = b , the line BB’ is called the **conjugate axis.**

∎ The latus rectum is the chord through a focus at right angle to the **Transverse axis.**

∎ The length of the semi-latus rectum obtained by putting x = ae in the equation of the hyperbola is

$ \displaystyle y = b\sqrt{\frac{a^2 e^2}{a^2} – 1} $

$ \displaystyle y = b\sqrt{e^2 – 1} = \frac{b^2}{a}$

### Relation Between Focal Distances :

The difference of the focal distances of a point on the hyperbola is constant. PM and PM’ are perpendiculars to the directrices MZ and M’Z’

PS’ − PS = e(PM’ − PM)

= eMM’ = e(2a/e)

= 2a = constant.

**Remark: **

∎ Using this property hyperbola can be defined in another way − Locus of a moving point such that difference of its distances from two fixed point is constant, would be hyperbola.

For example, interference fringes formed in Young’s Double Slit experiment are hyperbolic in nature.

### Relative Position of a Point with respect to the Hyperbola:

The quantity $ \displaystyle \frac{x_1^2}{a^2} – \frac{y_1^2}{b^2} = 1 $ is positive, zero or negative

i.e. ( S_{1} > = or < 0 ) , according as the point (x_{1 }, y_{1}) lies within , upon or without the curve, where

$ \displaystyle S_1 = \frac{x_1^2}{a^2} – \frac{y_1^2}{b^2} – 1 $

Note that this relation is converse in hyperbola if we compare with that in circle, parabola or ellipse.

__Parametric Coordinates:__

We can express the coordinates of a point of the hyperbola $ \displaystyle \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 $

in terms of a single parameter, say θ

In the adjacent figure OM = a secθ and PM = b tanθ.

Thus any point on the curve , in parametric form is x = a secθ, y = b tanθ

I**n other words ,** (a secθ, b tanθ) is a point on the hyperbola for all values of θ. The point (a secθ, b tanθ) is briefly written the point ‘ θ ‘

### Important Properties of Hyperbola:

Since the fundamental equation of the hyperbola only differs from that of the ellipse in having −b^{2 }instead of b^{2} , it will be found that many propositions for the hyperbola are derived from those for the ellipse by changing the sign of b^{2} .

**Some results for the hyperbola $ \displaystyle \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1 $
are :**

(i) The tangent at any point (x1, y1) on the curve is $ \displaystyle \frac{x x_1}{a^2} – \frac{y y_1}{b^2} = 1 $

(ii) The tangent at the point ‘ θ ‘ is $ \displaystyle \frac{x sec\theta}{a} – \frac{y tan\theta}{b } = 1 $

(iii) The straight line y = mx + c is a tangent to the curve , if c^{2} = a^{2} m^{2} − b^{2} .

In other words , $ \displaystyle y = m x \pm \sqrt{a^2 m^2 – b^2} $

touches the curve, for all those values of m when m > b/a or m < − b/a.

If m = ±b/a the tangents touch at infinity and are called as **asymptotes.**

(iv) Equation of the normal at any point (x1, y1) to the curve is $ \displaystyle \frac{x-x_1}{x_1^2 /a^2} = \frac{y-y_1}{y_1/(-b^2)} $

(v) The equation of the chord through the points θ_{1} and θ_{2} is

$ \large \left| \begin{array}{ccc} x & y & 1 \\ a sec\theta_1 & b tan\theta_1 & 1 \\ a sec\theta_2 & b tan\theta_2 & 1\end{array} \right| =0 $

(vi) The equation of the normal at θ is ax cosθ + by cotθ = a^{2} + b^{2}

(vii) Through a given point, four normals can be drawn to a hyperbola (real or imaginary).

(viii) Normal drawn at any point bisects the angle between the lines, joining the point to the foci, whereas the tangent at the same point bisects the supplementary angle between the lines.

(ix) Equation of director the circle is x^{2} + y^{2} = a^{2} – b^{2} .

That means if a^{2} > b^{2}, there would exist several points such that tangents drawn from them would be mutually perpendicular.

If a^{2} < b^{2} , no such point exist.

For a^{2} = b^{2} , center is the only point from which two perpendicular tangents (asymptotes) to the hyperbola can be drawn.

**Illustration : ** Find the equation to the hyperbola, whose eccentricity is 5/4 , where focus is (a, 0) and whose directrix is 4x − 3y = a. Also find the co−ordinates of the centre and the equation to the other directrix.

**Solution: **By definition, the equation to the hyperbola having the focus (a , 0) and directrix as 4x − 3y = a , and eccentricity 5/4 , will be

$ \displaystyle (x-a)^2 + (y-0)^2 = (\frac{5}{4})^2 [\frac{4x-3y-a}{\sqrt{4^2+ 3^2}}]^2 $

or (x^{2} − 2ax + a^{2} + y^{2}) 16 = 16x^{2} + 9y^{2} + a^{2} − 24xy − 8ax + 6ay

or 7y^{2} + 24xy − 24ax − 16ay + 15a^{2} = 0

**This is the equation to required hyperbola.**

Let the given focus (a, 0) be S. the axis will be a line passing through (a , 0) and perpendicular to the directrix

4x − 3y = a

Any line perpendicular to directrix is 3x + 4y = l

(where l is any constant)

As the line passes through (a, 0), we have

3a + 0 = λ

⇒ λ = 3a ,

Hence equation to the axis is 3x + 4y = 3a . . . (2)

Let Z be the point of intersection of the axis (2) with the directrix(1), then the co-ordinates of Z will be obtained by solving (2) and (1)

Solving we get the co-ordinates as (13a/25 , 9a/25)

The centre divides the join of S and Z externally in the ratio

1: e^{2} or 1 : (5/4)^{2} or 16 : 25.

Hence the co-ordinates of the centre C will be

$ \displaystyle (\frac{16\times a – (13a/25)\times 25}{16-25} , \frac{16 \times 0 -(25\times 9a)/25}{16-25})$

or -3a/9 and 9a/9 or ( -a/3 , a)

Let the other directix intersect the axis at Z’ and (x, y) be its co-ordinates, then

x + 13a/25 = −2a/3 and y + 9a/25 = 2a

or x = −89a / 75 and y = 41a/25

Therefore, the other directix is the line passing through the point ( −89a / 75 , 41a / 25) and parallel to the given directix (1) may be written as

4x − 3y = λ . . . (3)

As (3) passes through ( −89a / 75 , 41a/25) , we have λ = − 29a/3

Substituting the value of λ in (3)

we get the equation to the other directix as

4x − 3y = −29a/3

or 12x − 9y +29a = 0

**Illustration :** Prove that equation $ \displaystyle \sqrt{(x+4)^2 + (y+2)^2} – \sqrt{(x-4)^2 + (y-2)^2} = 8 $ represents a hyperbola.

**Solution:** [(x + 4)^{2} + (y + 2)^{2}] − [(x − 4)^{2} + (y − 2)^{2}] = 16 x + 8y ….(1)

$ \displaystyle \sqrt{(x+4)^2 + (y+2)^2} – \sqrt{(x-4)^2 + (y-2)^2} = 8 $ ….(2)

Dividing (1) by (2), we get

$ \displaystyle \sqrt{(x+4)^2 + (y+2)^2} + \sqrt{(x-4)^2 + (y-2)^2} = \frac{16x +8y}{8}$ …(3)

From (2) and (3),

$ \displaystyle 2 \sqrt{(x+4)^2 + (y+2)^2} = \frac{16x + 8y +64}{8} $

$ \displaystyle \sqrt{(x+4)^2 + (y+2)^2} = \frac{8x + 4y + 32}{8} $

$\large = \frac{\sqrt{5}}{2} (\frac{2x + y + 8}{\sqrt{5}})$

Clearly it is a hyperbola with eccentricity √5/2

**Illustration : ** Prove that the straight lines $\large \frac{x}{a}- \frac{y}{b} = m $ and $\large \frac{x}{a} + \frac{y}{b} = \frac{1}{m} $ , where a and b are given positive real numbers and ‘ m ‘ is a parameter , always meet on a hyperbola.

**Solution: **Here , $\large (\frac{x}{a}- \frac{y}{b})(\frac{x}{a} + \frac{y}{b}) = m \frac{1}{m} = 1$

∴ $\large \frac{x^2}{a^2}- \frac{y^2}{b^2} = 1 $

which is hyperbola.

**Illustration :** Let m_{1} and m_{2} be slopes of tangents from a point (1 , 4) on the hyperbola $\large \frac{x^2}{25}- \frac{y^2}{16} = 1 $

. Find the point from which the tangents drawn on the hyperbola have slopes |m_{1}| and |m_{2}| and positive intercepts on y−axis.

**Solution: **Any tangent to the hyperbola $\large \frac{x^2}{25}- \frac{y^2}{16} = 1 $ of slope m is

$\large y = m x \pm \sqrt{25m^2 – 16}$

⇒ (4 − m)^{2} = 25 m^{2} − 16

⇒ 3m^{2} + m − 4 = 0

⇒ m_{1} = 1 , m_{2} = − 4/3

⇒ |m_{1}| = 1 , m_{2} = |4/3|

Therefore, tangents are y = x + 3 and y = 4x/3 + 16/3

Solving them, we get (− 7 , −4) as the required point.

**Illustration :** The normal to the hyperbola meets the axes in M and N , and lines MP and NP are drawn at right angles at the axes, prove that the locus of P is hyperbola (a^{2} x^{2} − b^{2} y^{2}) = (a^{2} + b^{2})^{2}

**Solution:** Equation of normal at any point Q is ax cosθ + by cotθ = a^{2} + b^{2}

$\large M=(\frac{a^2 + b^2}{a} sec\theta , 0) , N=(0 , \frac{a^2 + b^2}{b} tan\theta)$

∴ Let P≡ (h , k)

$\large h = \frac{a^2 + b^2}{a} sec\theta , k = \frac{a^2 + b^2}{b} tan\theta$

sec^{2}θ − tan^{2}θ = 1

$\large \frac{a^2 h^2}{a^2 + b^2} – \frac{b^2 k^2}{a^2 + b^2} = 1 $

∴ locus of P is (a^{2}x^{2} − b^{2}y^{2}) = (a^{2} + b^{2})

**Illustration :** Show that the normal to the rectangular hyperbola xy = c^{2} at the point ‘ t ‘ meets the curve again at a point t’ such that t^{3} t’ = −1

**Solution:** Equation of normal at t is xt^{3} − yt − ct^{4} + c = 0

it passes through (ct’ , c/t’)

=> ct’t^{3} − c/t’t − ct^{4} + c = 0

=> (t^{3}t’ + 1) (t’− t) = 0

=> t^{3}t’ = −1 ( t’ − t ≠ 0)

**Exercise :**

(i) Find the centre, eccentricity, foci and directrices of the hyperbola

16x^{2} – 9y^{2} + 32x + 36y – 164 = 0.

(ii) Find the equation of the hyperbola whose one focus is (–1, 1), eccentricity = 3 and the equation of the corresponding directrix is x – y + 3 = 0.

(iii) A point moves such that difference of its distances from two given points is constant. Prove that the locus of the point is a hyperbola.

(iv) The hyperbola $\large \frac{x^2}{a^2}- \frac{y^2}{b^2} = 1 $ passes through the point of intersection of the lines 7x + 13y – 87 = 0 and 5x – 8y + 7 = 0 and its latus rectum is 32√2/5 . Find a and b.

(v) Find the locus of the points of intersection of two tangents to a hyperbola x²/25 −y²/16 = 1 if sum of their slopes is constant a.

(vi) Prove that the point of intersection of two tangents to the hyperbola $\large \frac{x^2}{a^2}- \frac{y^2}{b^2} = 1 $ , the product of whose slopes is c^{2}, lies on the curve y^{2} + b^{2} = c^{2}(x^{2} – a^{2}).

(vii) If the normal at the point Q on the hyperbola $\large \frac{x^2}{a^2}- \frac{y^2}{b^2} = 1 $ , whose vertices are A and A’ , meets the transverse axis at G, then prove that AG × A’G = a^{2}(e^{4}sec^{2θ} – 1).

### Also Read :

Standard Equation & Basic Definitions |