Question

3. Find the equation of the cubic which has the same asymptotes as those of the curve x-6x²y +11xy? -6y? + x + y + 1 = 0 and which passes through the points (0,0). (1.0) and (0,1).​

Answer 1

Correct Question: Find the equation of the cubic which has the same asymptotes as those of the curve $x^3 - 6x^2y + 11xy^2 - 6y^3 + x+y+1=0$

and which passes through the points (0,0) ,(1,0), (0,1).

Answer: $x^3 - 6x^2y + 11xy^2 - 6y^3 - x +6y =0$

Given:  Cubic equation passes through the points (0,0). (1.0) and (0,1).​

To find: Equation of cubic.

Step-by-step explanation:

Given equation

$x^3 - 6x^2y + 11xy^2 - 6y^3 + x+y+1=0$

or

(x-y)(x-2y)(x-3y) + (x+y+1) = 0

it is of the form

F3 + F1 = 0

Asymptotes are given by

F3 = 0

(x-y)(x-2y)(x-3y) = 0

or

x = y

x = 2y

x = 3y

Equation of cubic with asymptotes given by

F3 = 0

is F3 + F1 = 0

F1 = polynomial of degree 1

Equation of cubic is

(x-y)(x-2y)(x-3y) +ax + by + c = 0

where a,b,c are constants

Now the cube passes through the points

(0,0). (1.0) and (0,1).​

Putting values in the equation of cubic.

we get

c = 0

a = -1

b = 6

Hence cubic equation is given by

(x-y)(x-2y)(x-3y) - x + 6y  = 0

It can be written as

$x^3 - 6x^2y + 11xy^2 - 6y^3 - x +6y =0$

Answer: $x^3 - 6x^2y + 11xy^2 - 6y^3 - x +6y =0$

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