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

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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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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