The equation of the tangent to the hyperbola `3x^2-y^2=3` which is perpendicular to the line `x+3y-2=0` is

Options

A. `y=3x+-sqrt(6)`
B. `y=3x+-2sqrt(3)`
C. `y=3x+-sqrt(3)`
D. none of these

Explanation

Let `m` to be the slope of the tangent to the given hyperbola. Then,
`mxx("Slope of the line"x+3y=2)=-1`
`impliesmxx(-(1)/(3))=-1impliesm=3`
The equation of the hyperbola is
`3x^(2)-y^(2)=3implies(x^(2))/(1)-(y^(2))/(3)=1`
This is of the form `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, where `a^(2)=1` and `b^(2)=3`.
So, the equation of the tangents are
`y=mx+-sqrt(a^(2)m^(2)-b^(2))`
or, `y=3x+-sqrt(9-3)impliesy=3x+-sqrt(6)`