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Question
Find the equation of the tangent to the hyperbola:
3x2 – y2 = 4 at the point `(2, 2sqrt(2))`
Solution
The equation of the hyperbola is 3x2 – y2 = 4
i.e. `x^2/((4/3)) - y^2/4` = 1
Comparing with `x^2/"a"^2 - y^2/"b"^2` = 1, we get,
a2 = `4/3`, b2 = 4
The equation of the tangent to `x^2/"a"^2 - y^2/"b"^2` = 1 at the point (x1, y1) on it is
`("xx"_1)/"a"^2 = (yy_1)/"b"^2` = 1
∴ the equation of the tangent to the given hyperbola at the point `(2, 2sqrt(2))` is
`(x(2))/((4/3)) - (y(2sqrt(2)))/4` = 1
∴ `(3x)/2 - (sqrt(2)y)/2` = 1
∴ `3x - sqrt(2)y` = 2.
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