The equation of the tangent parallel to `y-x+5=0" drawan to "(x^(2))/(3)-(y^(2))/(2)=1` is

Options

A. `x-y-1=0`
B. `x-y+2=0`
C. `x+y-1=0`
D. `x+y+2=0`

Answer

A

Explanation

Given hyperbola is `(x^(2))/(3)-(y^(2))/(2)" "....(i)`
Equation of tangent parallel to `1y-x+5=0` is
`y-x+lambda=0`
`rArr y=x-lambda " "....(ii)`
If line (ii) is a tangent to hypervbola (i), thetn
`-lambda+pm sqrt(3x-2)`
(From `c+- sqrt(a^(2)m^(2)-b^(2)))`
`rArr -lambda+-1`
`rArr lambda=-1+1`
Piut the values of `lambda` in Eq. (ii) we get `x-y-1=0 and x-y+1=0` are the required tangents.