If the line `2x+sqrt(6)y=2` touches the hyperbola `x^2-2y^2=4` , then the point of contact is `(-2,sqrt(6))` (b) `(-5,2sqrt(6))` `(1/2,1/(sqrt(6)))` (d) `(4,-sqrt(6))`

Options

A. `(-2,sqrt(6))`
B. `(-5,2sqrt(6))`
C. `((1)/(2),(1)/(sqrt(6)))`
D. `(4,-sqrt(6))`

Answer

D

Explanation

The equation of the hyperbola is
`x^(2)-2y^(2)=4implies(x^(2))/(2^(2))-(y^(2))/(sqrt(2))^(2)=1`……..`(i)`
The equation of the line is
`2x+sqrt(6y)=2impliesy=-sqrt((2)/(3))x+(1)/(sqrt(6))`……….`(ii)`
If the line `y=mx+c` touches the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, then the coordinates of the point of contact are
`(+-(a^(2)m)/(sqrt(a^(2)m^(2)-b^(2))),overset(-)(+)(b^(2))/(sqrt(a^(2)m^(2)-b^(2))))`
Here, `a=2`, `b=sqrt(2)` and `m=-sqrt((2)/(3))`
So, the coordinates of the point of contact are `-=(overset(-)(+)4, +-sqrt(6))`.