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The coordinates of a point on the curve y = x ln (ex) at which the tangent is perpendicular to the line x + 2y + 3 = 0, are |
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Answer» Given curve is y = x ln (e x) \(\therefore\) slope of tangent to curve is \(\frac{dy}{dx}=\frac{x}{ex}\times e+ln(ex)\) = 1 + ln(ex) = lne + ln ex = ln(e2x) Given line is x + 2y + 3 = 0 slope of perpendicular line to line x + 2y + 3 = 0 = \(\frac{-1}{slope\,of\,line\,x + 2y+3=0}\) = \(\cfrac{-1}{\frac{-1}2}\) = 2 Since, given that tangent to curve is perpendicular to the line x + 2y + 3 = 0 So, their slopes are equal. \(\therefore\) ln(e2x) = 2 ⇒ e2x = e2 ⇒ x = 1 \(\therefore\) y = x ln(ex) = 1 ln(e) = 1 Hence, required point is (1, 1) which the tangent to given curve is perpendicular to the line x + 2y + 3 = 0 |
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