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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.

Standard Equation of an Hyperbola

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,

SAAK = e : 1      

⇒ SA = e  ∙ AK …………. (ii)

and  SAAK =  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 = ae ………………. (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

⇒ Sp2= e2PM2

⇒ SP2 = e2KN2

⇒ SP2 = e2(CN - CK)2

⇒ (x - ae)2 + y2 = e2(x - ae)2, [From (iii) and (iv)]

⇒ x2 - 2aex + (ae)2 + y2 = (ex - a)2

⇒ (ex)2 - 2aex + a2 = x2 - 2aex + (ae)2 + y2

⇒ (ex)2  - x2 - y2 = (ae)2 - a2

⇒ x2(e2 - 1) - y2 = a2(e2 - 1)

x2a2 - y2a2(e21) = 1

We know that a2(e2 - 1) = b2

Therefore, x2a2 - y2b2 = 1

For all the points P (x, y) the relation x2a2 - y2b2 = 1 satisfies on the required hyperbola.

Therefore, the equation x2a2 - y2b2 = 1 represents the equation of the hyperbola.

The equation of a hyperbola in the form of x2a2 - y2b2 = 1 is known as the standard equation of the hyperbola.

The Hyperbola





11 and 12 Grade Math 

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