Mathematics Asked by lkpppppppp on February 2, 2021
Find an equation relating $x$ and $y$ corresponding to the parametric equations
$$x=dfrac {3t}{(1+t^3)
}\y=dfrac {6t^2}{(1+t^3)}$$
Where $t ne −1$
by eliminating the variable $t$ and expressing the relation between $x$ and $y$ in the form
$P(x,y)=0$,
where $P(x,y)$ is a polynomial in $x$ and $y$ such that the coefficient of $x^3$ is $216$.
Answer:
eqt$=0$
What is the equation?
$$x=dfrac {3t}{(1+t^3) },y=dfrac {(6t^2)}{(1+t^3)}\frac yx=dfrac{dfrac {(6t^2)}{(1+t^3)}}{dfrac {3t}{(1+t^3) }}=2t$$ so you can find $t$ and put $t$ in one of relations
can you take over ?
final hint
$$t=frac y{2x}to \
x=dfrac {3t}{(1+t^3)}to x= dfrac {3frac y{2x}}{(1+(frac y{2x})^3)}$$
Answered by Khosrotash on February 2, 2021
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