Tangents which are parrallel to the line `2x+y+8=0` are drawn to hyperbola `x^(2)-y^(2)=3`. The points of contact of these tangents is/are

Options

A. (2,1)
B. `(2,-1)`
C. `(-2,-1)`
D. `(-2,1)`

Answer

B, D

Explanation

Equation of tangent parallel to the line `2x+y+8=0` is `y=-2x+c.`
Solving with hyperbola `x^(2)-y^(2)=3`, we get
`x^(2)-(-2x+c)^(2)=3`
Discriminant of this quadratic is zero.
`therefore" "c= pm 3`
So, equations of tangents are
`2x+y= pm3" (1)"`
Let `(x_(1),y_(1))` be the point of contact, then equation of tangent is
`x x_(1)-yy_(1)=3" (2)"`
Comparing Eqs. (1) and (2), we get
`(2)/(x_(1))=(1)/(-y_(1))= pm (3)/(3)`
`therefore" "x_(1)= pm and y_(1) = pm 1`
So, point of contact are `(2, -1) and (-2, 1)`.