Find the equation of the normal to the curve y=x+1/x, x>0 perpendicular to the line 3x-4y=7
Answers
Given info : The normal is normal to the curve is y = x + 1/x , x > 0 and normal perpendiculars to the line is 3x - 4y = 7.
To find : The equation of normal
solution : curve, y = x + 1/x
differentiating with respect to x,
dy/dx = 1 - 1/x²
slope of tangent of curve is dy/dx = 1 - 1/x²
a/c to question,
normal of the curve is perpendicular to the line 3x - 4y = 7 so tangent of the curve must be parallel to the line.
so, slope of tangent = slope of line
⇒1 - 1/x² = 3/4
⇒1 - 3/4 = 1/4 = 1/x²
⇒x = ±2
but x > 0 so x = 2
y = 2 + 1/2 = 5/2
now slope of normal = -1/slope of tangent = -1/(3/4) = -4/3
and now equation of normal is (y - 5/2) = -4/3(x - 2)
⇒(y - 5/2) + 4/3(x - 2) = 0
⇒3y - 15/2 + 4x - 8 = 0
⇒6y + 8x - 15 - 16 = 0
⇒8x + 6y - 31 = 0
Therefore the equation of normal is 8x + 6y - 31 = 0.