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Question:

If the tangent to the curve y=x/(x²−3), x∈R, (x≠±√3), at a point (α,β)≠(0,0) on it is parallel to the line 2x+6y−1=0, then:

|2α+6β|=19

|2α+6β|=11

|6α+2β|=19

|6α+2β|=9

Solution:

|dy/dx|(α,β)=−α/(2−3)(α²−3)²
Given that: −α/(2−3)(α²−3)²=−1/3
⇒α=0, ±√3
(α≠0) ⇒ β=±1/2
(β≠0)
|6α+2β|=19