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Question

If the tangent to the curve y=xx23,xR,(x±3,) at a point (α,β)(0,0) on it is parallel to the line 2x+6y11=0, then :

A
|6α+2β|=19
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B
|2α+6β|=11
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C
|6α+2β|=9
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D
|2α+6β|=19
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Solution

The correct option is A |6α+2β|=19
y=xx23 ...(1)
Differentiating w.r.t x, we get
dydx=x23(x23)2
dydx(α,β)=α23(α23)2

The slope of line is equal to dydx which is 13
α23(α23)2=13
α=0,± 3 (α0)

Now from equation (1)
β=αα23=±12

(α,β)=(3,12),(3,12)
|6α+2β|=19

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