Find the equation of a normal to the curve y = x loge x which is paral

1.	Find the equation of a normal to the curve y = x loge x which is parallel to the line 2x – 2y + 3 = 0.
Answer» finding the slope of the tangent by differentiating the curve \(\frac{dy}{dx}=Inx+1\) m(tangent) = In x + 1 normal is perpendicular to tangent so, m₁m₂ = – 1 m(normal) \(=-\frac{1}{Inx+1}\) equation of normal is given by y – y₁ = m(normal)(x – x₁) now comparing the slope of normal with the given equation m(normal) = 1 \(-\frac{1}{Inx+1}=1\) \(x=\frac{1}{e^2}\) since this point lies on the curve, we can find y by substituting x \(y=-\frac{2}{e^2}\) The equation of normal is given by \(y+\frac{2}{e^2}=x-\frac{1}{e^2}\)

Find the equation of a normal to the curve y = x loge x which is parallel to the line 2x – 2y + 3 = 0.

Answer»

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=Inx+1\)

m(tangent) = In x + 1

normal is perpendicular to tangent so, m₁m₂ = – 1

m(normal) \(=-\frac{1}{Inx+1}\)

equation of normal is given by y – y₁ = m(normal)(x – x₁)

now comparing the slope of normal with the given equation

m(normal) = 1

\(-\frac{1}{Inx+1}=1\)

\(x=\frac{1}{e^2}\)

since this point lies on the curve, we can find y by substituting x

\(y=-\frac{2}{e^2}\)

The equation of normal is given by

\(y+\frac{2}{e^2}=x-\frac{1}{e^2}\)