If the tangent to the curve `y= (x)/(x^(2) - 3), x in r ( xne pm sqrt3)`, at a point `(alpha, beta ) ne (0, 0) ` on it is parallel to the line `2x + 6y -11 =0`, then (a) |6α + 2β| = 9 (b) |2α + 6β| = 11 (c) |2α + 6β| = 19 (d) |6α + 2β| = 19

Answer

9

Explanation

`(dy)/(dx) = ((x^(2) -3) -x(2x))/((x^(2) - 3)^(2))`
`:'` tangent is parallel to line
`(-x^(2) -3)/((x^(2) -3)^(2)) = -(1)/(3) :. (x^(2) + 3)/((x^(2) - 3)^(2)) = (1)/(3)`
Put `x^(2) = t`
`3(t+ 3) = t^(2) + 9 - 6t rArr 3t + 9 = t^(2) + 9 - 6t rArr t^(2) -9t = 0`
`t = 0, 9`
`:. x^(2) = 0, 9 rArr x = +- 3`
for `x = 3, y = (3)/(6) = (1)/(2)`
for `x = -3, y = (-3)/(6) = (-1)/(2)`
Which satisfies option (2)