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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      

β‡’ SA = e  βˆ™ AK …………. (ii)

and  SAβ€²Aβ€²K =  e : 1    

β‡’ SA' = e  βˆ™ 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)

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

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

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

β‡’ CS - a + CS + a = e (a - CK + a + CK)

β‡’2CS = e (2a)

β‡’ 2CS = 2ae

β‡’ CS = ae …………………… (iii)

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

β‡’ SA' - SA = e (A'K - AK)

β‡’ AA'= e {(CA’ + CK) - (CA - CK)}

β‡’ AA' = e (CA’ + CK - CA + CK)

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

β‡’ AA' = e (a + CK - a + CK)

β‡’ 2a = e (2CK)

β‡’ 2a = 2e (CK)

β‡’ a = e (CK)

β‡’ 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 βˆ™ PM

β‡’ Sp2= e2PM2

β‡’ SP2 = e2KN2

β‡’ SP2 = e2(CN - CK)2

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

β‡’ x2 - 2aex + (ae)2 + y2 = (ex - a)2

β‡’ (ex)2 - 2aex + a2 = x2 - 2aex + (ae)2 + y2

β‡’ (ex)2  - x2 - y2 = (ae)2 - a2

β‡’ x2(e2 - 1) - y2 = a2(e2 - 1)

β‡’ x2a2 - y2a2(e2βˆ’1) = 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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