# Standard Equation of an Hyperbola

We will learn how to find the standard equation of a hyperbola.

Let S be the focus, e (> 1) be the eccentricity and line KZ its directrix of the hyperbola whose equation is required.

From the point S draw SK perpendicular to the directrix KZ. The line segment SK and the produced SK divides internally at A and externally at A’ respectively in the ratio e : 1.

Then,

$$\frac{SA}{AK}$$ = e : 1

⇒ SA = e  ∙ AK …………. (ii)

and  $$\frac{SA'}{A'K}$$ =  e : 1

⇒ SA' = e  ∙ A'K …………………. (ii)

The points A and A' he on the required hyperbola because according to the definition of hyperbola A and A’are such points that their distance from the focus bear constant ratio e (>1) to their respective distance from the directrix, therefore A and A' he on the required hyperbola.

Let AA’ = 2a and C be the mid-point of the line segment AA'. Therefore, CA = CA' = a.

Now draw CY perpendicular to AA’ and mark the origin at C. CX and CY are assumed as x and y-axes respectively.

Now, adding the above two equations (i) and (ii) we have,

SA + SA' = e (AK + A'K)

⇒ CS - CA + CS + CA' =  e (AC - CK + A’C + CK)

⇒ CS - CA + CS + CA' =  e (AC - CK + A’C + CK)

Now put the value of CA = CA' = a.

⇒ CS - a + CS + a = e (a - CK + a + CK)

⇒2CS = e (2a)

⇒ 2CS = 2ae

⇒ CS = ae …………………… (iii)

Now, again subtracting above two equations (i) from (ii) we have,

⇒ SA' - SA = e (A'K - AK)

⇒ AA'= e {(CA’ + CK) - (CA - CK)}

⇒ AA' = e (CA’ + CK - CA + CK)

Now put the value of CA = CA' = a.

⇒ AA' = e (a + CK - a + CK)

⇒ 2a = e (2CK)

⇒ 2a = 2e (CK)

⇒ a = e (CK)

⇒ CK = $$\frac{a}{e}$$ ………………. (iv)

Let P (x, y) be any point on the required hyperbola and from P draw PM and PN perpendicular to KZ and KX respectively. Now join SP.

According to the graph, CN = x and PN = y.

Now form the definition of hyperbola we get,

SP = e PM

⇒ Sp$$^{2}$$= e$$^{2}$$PM$$^{2}$$

⇒ SP$$^{2}$$ = e$$^{2}$$KN$$^{2}$$

⇒ SP$$^{2}$$ = e$$^{2}$$(CN - CK)$$^{2}$$

⇒ (x - ae)$$^{2}$$ + y$$^{2}$$ = e$$^{2}$$(x - $$\frac{a}{e}$$)$$^{2}$$, [From (iii) and (iv)]

⇒ x$$^{2}$$ - 2aex + (ae)$$^{2}$$ + y$$^{2}$$ = (ex - a)$$^{2}$$

⇒ (ex)$$^{2}$$ - 2aex + a$$^{2}$$ = x$$^{2}$$ - 2aex + (ae)$$^{2}$$ + y$$^{2}$$

⇒ (ex)$$^{2}$$  - x$$^{2}$$ - y$$^{2}$$ = (ae)$$^{2}$$ - a$$^{2}$$

⇒ x$$^{2}$$(e$$^{2}$$ - 1) - y$$^{2}$$ = a$$^{2}$$(e$$^{2}$$ - 1)

⇒ $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{a^{2}(e^{2} - 1)}$$ = 1

We know that a$$^{2}$$(e$$^{2}$$ - 1) = b$$^{2}$$

Therefore, $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1

For all the points P (x, y) the relation $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 satisfies on the required hyperbola.

Therefore, the equation $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 represents the equation of the hyperbola.

The equation of a hyperbola in the form of $$\frac{x^{2}}{a^{2}}$$ - $$\frac{y^{2}}{b^{2}}$$ = 1 is known as the standard equation of the hyperbola.

The Hyperbola

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.

## Recent Articles

1. ### Adding 1-Digit Number | Understand the Concept one Digit Number

Sep 17, 24 02:25 AM

Understand the concept of adding 1-digit number with the help of objects as well as numbers.

2. ### Counting Before, After and Between Numbers up to 10 | Number Counting

Sep 17, 24 01:47 AM

Counting before, after and between numbers up to 10 improves the child’s counting skills.

3. ### Worksheet on Three-digit Numbers | Write the Missing Numbers | Pattern

Sep 17, 24 12:10 AM

Practice the questions given in worksheet on three-digit numbers. The questions are based on writing the missing number in the correct order, patterns, 3-digit number in words, number names in figures…

4. ### Arranging Numbers | Ascending Order | Descending Order |Compare Digits

Sep 16, 24 11:24 PM

We know, while arranging numbers from the smallest number to the largest number, then the numbers are arranged in ascending order. Vice-versa while arranging numbers from the largest number to the sma…