Q. 194.5( 2 Votes )

The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 is
A. 3 (x – 6)2 – (y – 2)2 = 3

B. (x – 6)2 – 3 (y – 2)2 = 1

C. (x – 6)2 – 2(y – 2)2 = 1

D. 2(x – 6)2 – (y – 2)2 = 1

Answer :

Given: Foci is (4, 2), e = 2 and center at (6, 2)


To find: equation of the hyperbola


Formula used:


Standard form of the equation of hyperbola is,



Center is the mid-point of two vertices


The distance between two vertices is 2a


The distance between the foci and vertex is ae – a and b2 = a2(e2 – 1)


The distance between two points (m, n) and (a, b) is given by



Mid-point theorem:


Mid-point of two points (m, n) and (a, b) is given by



Therefore


Let one of the two foci is (m, n) and the other one is (4, 2)


Since, Centre(6, 2)





Foci are (4, 2) and (8, 2)


The distance between the foci is 2ae and Foci are (4, 2) and (8, 2)











a2 = 1


b2 = a2(e2 – 1)



b2 = 4 – 1


b2 = 3


Equation of hyperbola:





3(x – 6)2 – (y – 2)2 = 3


Hence, required equation of hyperbola is 3(x – 6)2 – (y – 2)2 = 3

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