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Mathematics

2 answers:

Morgarella [4.7K]4 years ago

7 0

Answer:

Option d)

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

Step-by-step explanation:

we are given with the equation

$3x^2+y^2=9$

Dividing both sides by 9 and simplifying we get

$\frac{x^2}{3}+\frac{y^2}{9}=1$

$\frac{x^2}{(\sqrt{3})^{2}}+\frac{y^2}{3^{2}}=1$

Which represents an ellipse. Hence we have an ellipse in our problem. And also it is obvious that any translation or rotation of this ellipse will again result into en ellipse.

If we see the first two options , they are the equations of the parabolas hence they can not be answer to the problem.

Let us see the third equation.

It is

$3x^2+y^2+3x-3y+3=0$

Let us transform this equation in perfect square form.

$3x^2+3x+y^2-3y+3=0$

$3(x^2+x)+y^2-3y+\frac{9}{4}-\frac{9}{4}+3=0$

$3(x^2+x+\frac{1}{4}-\frac{1}{4})+y^2-3y+\frac{9}{4}-\frac{9}{4}+3=0$

$3(x^2+x+\frac{1}{4})-3\times\frac{1}{4}+y^2-3y+\frac{9}{4}-\frac{9}{4}+3=0$

$3(x+\frac{1}{2})^{2}+(y+\frac{3}{2})^{2}-3\times\frac{1}{4}-\frac{9}{4}+3=0$

$3(x+\frac{1}{2})^{2}+(y+\frac{3}{2})^{2}=0$

Which represents the equation of an circle , Although Circle also comes in the category of an ellipse but it clearly not the same as we have in the problem in which the minor and major axis are different.

Hence this is also not our answer. So we have a clue that as the first three options are not correct , the right answer must be the forth one. Let us confirm it by converting it into a perfect square.

The equation given is

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

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