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If $x,y \in \mathbb{R}.$ Then the equation

$3x^4-2(19y+8)x^2+361y^2+2(100+y^4)+64=2(190y+2y^2)$ represent in rectangular cartesian system

Options

$(a)$ Circle $\;\;\;\;\;(b)$ Parabola $\;\;\;\;\;\; (c)$ Ellipse $\;\;(d)$ Hyperbola

Try: From $$3x^4-2(19y+8)x^2++2y^4+357y^2-380y+264=0$$

For real roots, its discriminant always $\geq 0$

$$4(19y+8)^2-4\cdot 3 \cdot (2y^4+357y^2-380y+264)\geq 0$$

$$361y^2+64+304y-6y^4-1125y^2+1140y-792\geq 0$$

So $$6y^4+764y^2-1444y+728\leq 0$$

I am struck at that point. did not how to solve further

$\bf{Added}:$ I have seems that it can be convert into sum of square of numbers

like $()^2+()^2+()^2+\cdots =0$

Could some help me to convert it , thanks

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1 Answers1

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Points $(3,1)$ and $(-3,1)$ belong to the curve. On the other hand, the equation you get letting $x=0$ has no real solutions, i.e. the curve doesn't cross the y-axis and is thus not connected.

Among the proposed alternatives, only the hyperbola is not connected, hence it must be the solution.

EDIT.

It turns out, however, that this equation cannot represent a conic, as it cannot be factored, hence the question is ill-posed. As mathlove points out in the comment below, $(3,1)$ and $(-3,1)$ are its only real solutions.

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