Question

find the asymptote of the curve x^2(x+y)(x-y)+2x^3(x-y)-4y^3=0​

Answer 1

[tex]

Given :Equation :(2x+3)y=(x-1)^2(2x+3)y=(x−1)

2

To find :Find the asymptotes of the curve

Solution:

We are supposed to find the asymptotes of the curve

Equation :(2x+3)y=(x-1)^2(2x+3)y=(x−1)

2

\begin{gathered}\Rightarrow y =\frac{(x-1)^2}{(2x+3)}\\\Rightarrow y = \frac{x^2-2x+1}{2x+3}\end{gathered}

⇒y=

(2x+3)

(x−1)

2

⇒y=

2x+3

x

2

−2x+1

Vertical asymptote : Vertical asymptotes are vertical lines which correspond to the zeroes of the denominator of a rational function.

Equate numerator of given curve equals to 0

So,2x+3=0

x=\frac{-3}{2}x=

2

−3

So, Vertical asmptote is -2/3

Horizontal asymptote :A horizontal asymptote is a y-value on a graph which a function approaches but does not actually reach.

The given polynomial in the numerator is a higher degree than the denominator, there is no horizontal asymptote.

Slant asymptote:A slant (oblique) asymptote occurs when the polynomial in the numerator is a higher degree than the polynomial in the denominator.

To find the slant asymptote you must divide the numerator by the denominator

On dividing numerator by denomintaor

Quotient :\frac{x}{2}-\frac{7}{4}

2

x

−

4

7

So, Oblique asymptote :y =\frac{x}{2}-\frac{7}{4}y=

2

x

7

Answer 2

Concept Introduction: Curved Graphs are advanced concepts of Mathematics.

Given:

We have been Given:

${x}^{2}(x + y)(x - y) + 2 {x}^{3}(x - y) - 4 {y}^{3} = 0$

To Find:

We have to Find: find the asymptote of the curve.

Solution:

According to the problem, on simplifying the equation we get,

$3 {x}^{4} - {x}^{2} {y}^{2} - 2 {x}^{3}y - 4 {y}^{3} = 0$

Now sorting out, we get,

$3 {x}^{4} - 2 {x}^{3} y = 4 {y}^{3} + {x}^{2} {y}^{2}$

Now taking common outs we get our asymptote to be,

${x}^{3} (3x - 2y) = {y}^{2} (4y + {x}^{2} ) \\ \frac{ {x}^{3} }{ {y}^{2} } = \frac{ {x}^{2} + 4y}{3x - 2y}$

Final Answer: The Asymptote of the given Equation is,

$\frac{ {x}^{3} }{ {y}^{2} } = \frac{ {x}^{2} + 4y }{3x - 2y}$

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