Find the equation of hyperbola in each of the following cases:
(i) Centre is (1, 0), one focus is (6, 0) and transverse axis 6
(ii) Centre is (3, 2), one focus is (5, 2) and one vertex is (4, 2)
(iii) Centre is `(-3,2),` one vertex is `(-3,4)` and eccentricity is 5/2
(iv) Foci are (4,2), (8,2) and eccentricity is 2

Explanation

Centre is (1, 0), one focus is (6, 0).
So, trnsverse axis is along x-axis.
Therefore, equation of hyperbola is `((x-1)^(2))/(a^(2))-(y^(2))/(b^(2))=1.`
Given that 2a = 6.
`therefore" "a = 3`
Also, distance between centre and focus is 'ae'.
`therefore" "ae=6-1=5`
So, `b^(2)-a^(2)e^(2)-a^(2)=25-9=16`
Therefore, equation of hyperbola is `((x-1)^(2))/(9)-(y^(2))/(16)=1.`
(ii) Centre is (3, 2) and one focus is (5, 2).
Thus, transverse axis is horizontal , along line y = 2.
Therefore, equation of hyperbola is `((x-3)^(2))/(a^(2))-((y-2)^(2))/(b^(2))=1.`
Distance between centre and focus is 'ae'.
`therefore" "ae=5-3=2`
Also, one vertex is (4, 2).
`therefore" "a=4-3=1`
So, `b^(2)=a^(2)e^(2)-a^(2)=4-1=3`
Thereforem equation of hyperbola is `(x-3)^(2)-((y-2)^(2))/(3)`
= 1.
(iii) Centre is `(-3, 2)` and one vertex is `(-3, 4)`.
So, transverse axis is vertical, along line `x=-3`.
Thus, equation of hyperbola is `((x+3)^(2))/(a^(2))-((y-2)^(2))/(b^(2))=-1.`
Distance between centre and vertex is 'a'/
`therefore" "a=4-2=2`
Given that, e = 5/2
`therefore" "b^(2)=a^(2)e^(2)-a^(2)=25-4=21`
Therefore, equation of hyperbola is `((x+3)^(2))/(4)-((y-2)^(2))/(21)`
`=-1.`
Foci are (4, 2) and (8, 2)
So, transverse axis is horizontal, along line y = 2.
Also, centre is mid-point of foci, which is (6, 2).
Thus, equation of hyperbola is `((x-6)^(2))/(a^(2))-((y-2)^(2))/(b^(2))=1`.
Distance between foci, 2ae=4.
Also, it is given that e = 2.
`therefore" "a = 1`
So, `b^(2)=a^(2)e^(2)-1=4-1=3`
Therefore, equation of hyperbola is `(x-6)^(2)-((y-2)^(2))/(3)=1.`