Question

Find the equation of tangent at (3, 3) to the curve x³ + y³ - 6xy = 0 

Select one: 

 a. y = -x - 6 

 b. y = -x + 6 

 c. None of these answers 

 d. y = x + 6

 e. y = x - 6

Answer 1

Sol Biuen pont P (2,4)=(3,3) 23+ y ² bay=0 Ginen Curne Differentiate wor.to a 34²+ 3yrdy 6(1+y)=0 du art 42 dy a 22 dy - 2y = (9=20) = 24-24 dy ch 24-92 429 The slope of tangent to the came at pa 263) -C372 6-9 3 3 -- m 82_203) 9-6 et hom ... The equaton of tangent to the curre of curue at pu 4-4 = m 12-13 y-3 = -1 (1-3) Y-3 = -1+3 y = -2+3+3 y=-346 Ano: b

Answer 2

https://drive.google.com/file/d/1eksl02hvEVG1aCE7oWOE9SEuLGWcQ5gq/view?usp=drivesdk

Answer 3

SOLUTION :

Curve : x^3 + y^3 - 6 xy = 0

Differentiate w.r.to x :

=> 3x^2 + 3y^2 dy/dx - 6(x dy/dx + y) = 0

=> dy/dx (3y^2 - 6x) = (- 3x^2 +6y)

=> dy/dx = (6y - 3x^2) / (3y^2 - 6x) slope of tangent at point(x, y)

So, slope of tangent at point (3, 3) of the curve 

= (6(3) - 3(3)^2) / (3(3)^2 - 6(3))

= - 9/9

= - 1

Equation of this tangent through point (3, 3) :

=> (y - 3) = - 1(x - 3) (point-slope form (y- y1) = m(x - x1))

=> y - 3 = - x + 3

=> y = - x + 6 : Option b (ANSWER)