Question

Find the equation of a normal to the curve y=xlogx

Answer 1

equation of normal is x - y - 3/e² = 0

find the equation of a normal to the curve y=xlogx which is parallel to the line 2x-2y+3=0.

first we should find out slope of normal.

differentiating y = x logx with respect to x,

dy/dx = x × 1/x + logx = 1 + logx

we know, dy/dx is the slope of tangent of the curve y = f(x)

so, slope of normal = -1/(slope of normal ) [ normal and tangent are perpendicular to each other ]

so, slope of normal = -1/(1 + logx)

it is given, equation of line : 2x - 2y + 3 = 0

slope of line = 1

since line is parallel to normal

so slope of line = slope of normal

⇒1 = -1/(1 + logx)

⇒1 + logx = -1

⇒logx = -2, x = 1/e²

so, y = (1/e²)log(1/e²) = -2/e²

so equation of normal, (y + 2/e²) = 1(x - 1/e²)

⇒x - y - 3/e² = 0

also read similar questions: integration of 1/xlogx

https://brainly.in/question/1368624

xlogx+4 = 32, where base of logarithm is 2.

https://brainly.in/question/10267915

Answer 2

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find the equation of a normal to the curve y=xlogx which is parallel to the line 2x-2y+3=0.

first we should find out slope of normal.

differentiating y = x logx with respect to x,

dy/dx = x × 1/x + logx = 1 + logx

we know, dy/dx is the slope of tangent of the curve y = f(x)

so, slope of normal = -1/(slope of normal ) [ normal and tangent are perpendicular to each other ]

so, slope of normal = -1/(1 + logx)

it is given, equation of line : 2x - 2y + 3 = 0

slope of line = 1

since line is parallel to normal

so slope of line = slope of normal

⇒1 = -1/(1 + logx)

⇒1 + logx = -1

⇒logx = -2, x = 1/e²

so, y = (1/e²)log(1/e²) = -2/e²

so equation of normal, (y + 2/e²) = 1(x - 1/e²)

⇒x - y - 3/e² = 0