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A curve f(x) has normal at the point P(1,1) given by a(y – 1) +(x – 1) = 0. If the slope of tangent at any point on the curve is proportional to the ordinate at that point, then the curve is(a) y = eax–1(b) y –1 = eax(c) y = ea(x–1) (d) None of these |
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Answer» Correct option (c) y = ea(x – 1) Explanation: Normal at point P is ay + x = a + 1 Slope of tangent at P = a = (dy/dx) (1, 1) ...(1) Now dy/dx α ⇒ dy/dx = ky ⇒ (dy/dx)(1,1) = k ...(2) From (1) & (2) k = a dy/dx = ay ⇒ dy/y = a.dx ⇒ loge y = ax+C Now C = – a (as curve passes through (1,1)) loge y = ax–a ⇒ y = ea(x–1) |
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