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Show that 3x^(2) - 3y^(2) - 18x + 12y + ...

Show that `3x^(2) - 3y^(2) - 18x + 12y + 2 = 0` represents a rectangular hyperbola. Find its centre foci and eccentricity.

Text Solution

Verified by Experts

The given equation is
`3x^(2) - 3y^(2) -18x + 12y + 2 = 0`
`rArr 3(x^(2) -6x) -3(y^(2) -4y) + 2 = 0`
`rArr 3(x^(2) -6x + 9) -3(y^(2) -4y + 4) =13`
`rArr 3(x^(2) -6x +9) -3(y^(2) -4y + 4) =13`
`rArr ((x-3)^(2))/((13)/(3))-((y-2)^(2))/((13)/(3))=1`
Which represents a hyperbola in which the length of its transverse axis = length of conjugate axis
`=sqrt((13)/(3))`
Hence the given hyperbola is a rectangular hyperbola.
The centre of the hyperbola = (3, 2) and its eccentricity `= sqrt(2)`
The foci are `(3 pm sqrt((26)/(2)), 2)`.
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