If X and Y are given as current co-ordinates, what is the equation of the tangent at a specific point of x^3 – 3axy + y^3 = 0 at (x, y)?

Category: QuestionsIf X and Y are given as current co-ordinates, what is the equation of the tangent at a specific point of x^3 – 3axy + y^3 = 0 at (x, y)?
Editor">Editor Staff asked 11 months ago

If X and Y are given as current co-ordinates, what is the equation of the tangent at a specific point of x^3 – 3axy + y^3 = 0 at (x, y)?
 
(a) (x^2 – ay)X + (y^2 – ax)Y = -2axy
 
(b) (x^2 – ay)X + (y^2 – ax)Y = 2axy
 
(c) (x^2 – ay)X + (y^2 – ax)Y = axy
 
(d) (x^2 – ay)X + (y^2 – ax)Y = -axy
 
The question was asked during an interview for a job.
 
My doubt is from Calculus Application topic in chapter Application of Calculus of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

Right option is (c) (x^2 – ay)X + (y^2 – ax)Y = axy
 
Explanation: Equation of the given curve is, x^3 – 3axy + y^3 = 0 ……….(1)
 
Differentiating both sides with respect to x we get,
 
3x^2 – 3a(x(dy/dx) + y) + 3y^2(dy/dx) = 0
 
Or dy/dx = (ay – x^2)/(y^2 – ax)
 
So, it is clear that this can be written as,
 
Y – y = (dy/dx)(X – x)
 
Or Y – y = [(ay – x^2)/(y^2 – ax)](X – x)
 
Simplifying the above equation by cross multiplication, we get,
 
(x^2 – ay)X + (y^2 – ax)Y = x^3 – 3axy + y^3 + axy
 
Using (1),
 
(x^2 – ay)X + (y^2 – ax)Y = axy