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23 June, 14:32

Normal line to the curve y=ln (2x+a) at x=1 is perpendicular to x-4y+2 = 4ln3. Find the value of a.

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  1. 23 June, 16:31
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    Hello,

    The normal perpendicular to x-4y+2=4ln (3) or y = (x+2-4ln (3)) / 4

    has as slope - 4

    So y' = (ln (2x+a)) '=1 / (2x+a) * 2 if x=1, y'=1 / (2+a)

    Thus

    1 / (2+a) = - 4 = = >2+a=-1/4==>a=-1/4-2==>a=-9/4
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