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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 Aβ respectively in the ratio e : 1.
Then,
SAAK = e : 1
β SA = e β AK β¦β¦β¦β¦. (ii)
and SAβ²Aβ²K = e : 1
β SA' = e β A'K β¦β¦β¦β¦β¦β¦β¦. (ii)
The points A and A' he on the required hyperbola because
according to the definition of hyperbola A and Aβare 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 AAβ = 2a and C be the mid-point of the line segment AA'. Therefore, CA = CA' = a.
Now draw CY perpendicular to AAβ 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 + AβC + CK)
β CS - CA + CS + CA' = e (AC - CK + AβC + CK)
Now put the value of CA = CA' = a.
β CS - a + CS + a = e (a - CK + a + CK)
β2CS = e (2a)
β 2CS = 2ae
β CS = ae β¦β¦β¦β¦β¦β¦β¦β¦ (iii)
Now, again subtracting above two equations (i) from (ii) we have,
β SA' - SA = e (A'K - AK)
β AA'= e {(CAβ + CK) - (CA - CK)}
β AA' = e (CAβ + 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 β¦β¦β¦β¦β¦β¦. (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 β 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(e2β1) = 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.
β The Hyperbola
11 and 12 Grade Math
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