# The equation of the hyperbola whose centre is (6, 2) one focus is (4, 2) and of eccentricity 2 isA. 3 (x – 6)2 – (y – 2)2 = 3B. (x – 6)2 – 3 (y – 2)2 = 1C. (x – 6)2 – 2(y – 2)2 = 1D. 2(x – 6)2 – (y – 2)2 = 1

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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