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If the line 2x+sqrt(6)y=2 touches the hy...

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))`

A

`(-2,sqrt(6))`

B

`(-5, 2 sqrt(6))`

C

`((1)/(2),(1)/(sqrt(6)))`

D

`(4, -sqrt(6))`

Text Solution

Verified by Experts

The equation of tangent at `(x_(1),y_(1)) " is " x x_(1)-2yy_(1)=4,`
which is same as `2x+sqrt(6)y=2.`
`therefore (x_(1))/(2)=-(2y_(1))/(sqrt(6))=(4)/(2)`
`rArr x_(1)=4 and y_(1)= -sqrt(6)`
Thus, the point of contact is `(4, -sqrt(6)).`
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