Question

Find the equation of normal to the curve x^3+y^3=8xy at the point where it meet the curve y^2=4x other than the origin

Answer 1

solving equations : x³ + y³ = 8xy and y² = 4x

(y²/4)³ + y³ = 8(y²/4)y

or, y^6/64 + y³ = 2y³

or, y³/64 + 1 = 2

or, y³ = 64 => y = 4

and then x = y²/4 = 4²/4 = 4

so, point is (4, 4)

now find slope of tangent of curve x³ + y³ = 8xy at (4,4)

differentiating both sides,

3x² + 3y²y' = 8y + 8xy'

or, y' = (3x² - 8y)/(8x - 3y²)

putting x = 4 and y = 4

so, y' = (3 × 16 - 8 × 4)/(8 × 4 - 3 × 16)

= (48 - 32)/(32 - 48)

= -1

so, slope of normal = -1/slope of tangent

[ as we know tangent is perpendicular upon normal ]

so, slope of normal = 1

now equation of normal to the curve ;

(y - 4) = 1(x - 4)

or, x = y or, (x - y) = 0