Question

Identify the conic basic of x^2+xy+y^2+2x= -3

Answer 1

Explanation:

The given equation is:

x^2 + xy + y^2 + 2x = -3

To identify the conic basic of the equation, we need to check its discriminant, which is given by:

Δ = B^2 - 4AC

where A, B, and C are the coefficients of x^2, xy, and y^2 terms respectively.

In this case, A = 1, B = 1, and C = 1.

So,

Δ = B^2 - 4AC

= 1^2 - 4(1)(1)

= -3

Since the discriminant is negative, the given equation represents an ellipse.

Answer 2

The conic section basis of the equation x^2 + xy + y^2 + 2x = -3 can be determined by examining the coefficients of the x^2, xy, and y^2 terms.

To do this, we can start by completing the square for the quadratic terms in the equation:

x^2 + xy + y^2 + 2x = -3

(x^2 + 2x) + xy + y^2 = -3

(x + 1)^2 - 1 + xy + y^2 = -3

(x + 1)^2 + xy + y^2 = -2

Now, we can see that the coefficient of the xy term is positive, which indicates that the conic section is an ellipse. Specifically, this is a rotated ellipse because the x and y terms are not squared separately and have a non-zero coefficient.

Therefore, the conic section basis of the equation x^2 + xy + y^2 + 2x = -3 is a rotated ellipse.

Explanation: