Conic Sections Calculator

Analyze the equation $ x^2+y^2-25=0 $ and determine the type of conic it represents.

Equation analysis

Classification: Circle.

Standard form: $ x^2 + y^2 = 25 $

Step-by-step solution

1. Analysis of the conic type.

The given equation is:

$$ x^2 + y^2 - 25 = 0 $$

The general equation of a conic is:

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

By comparing, we extract the coefficients:

$$ A = 1 \quad B = 0 \quad C = 1 \\[0.5em] D = 0 \quad E = 0 \quad F = -25 $$

We use the discriminant to analyze the conic type:

$$ \Delta = B^2 - 4AC \\[0.5em] \Delta = 0 - 4(1)(1) = -4 $$

Since Δ < 0, the equation corresponds to a general ellipse.

2. Determinant analysis.

To check if the equation corresponds to a degenerate case, we calculate the following determinant:

$$ \delta = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix} \\[0.5em] \delta = 8ACF - 2AE^2 - 2B^2F + 2BDE - 2CD^2 \\[0.5em] \delta = 8(1)(1)(-25) - 2(1)(0)^2 - 2(0)^2(-25) + 2(0)(0)(0) - 2(1)(0)^2 \\[0.5em] \delta = -200 $$

Since δ ≠ 0, the equation does not correspond to a degenerate case.

3. Existence analysis.

To check if the equation corresponds to a real or imaginary ellipse, we use the trace invariant:

$$ A + C = 1 + 1 = 2 $$

Since the sign of A + C is different from the sign of δ, the equation corresponds to a real ellipse. Additionally, since A = C and B = 0, the conic is a circle.

Graph in the Cartesian plane of a circle centered at the origin obtained from the analysis of a general quadratic equation. — Graph of the circle

Determine the type of conic represented by the equation $ 9x^2+16y^2-144=0. $

General equation analysis

Classification: Non-degenerate ellipse.

Orientation: Horizontal (principal axis parallel to the x-axis).

Standard form: $ \dfrac{x^2}{16} + \dfrac{y^2}{9} = 1 $

For more information about the conic, check the ellipse calculator.

Step-by-step solution

1. Analysis of the conic type.

The given equation is:

$$ 9x^2 + 16y^2 - 144 = 0 $$

The general equation of a conic is:

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

By comparing, we extract the coefficients:

$$ A = 9 \quad B = 0 \quad C = 16 \\[0.5em] D = 0 \quad E = 0 \quad F = -144 $$

We will perform an invariant analysis to determine the conic type.

We use the discriminant to analyze the conic type:

$$ \Delta = B^2 - 4AC \\[0.5em] \Delta = 0 - 4(9)(16) = -576 $$

Since Δ < 0, the equation corresponds to a general ellipse.

2. Determinant analysis.

To check if the equation corresponds to a degenerate case, we calculate the following determinant:

$$ \delta = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix} \\[0.5em] \delta = 8ACF - 2AE^2 - 2B^2F + 2BDE - 2CD^2 \\[0.5em] \delta = 8(9)(16)(-144) - 2(9)(0)^2 - 2(0)^2(-144) + 2(0)(0)(0) - 2(16)(0)^2 \\[0.5em] \delta = -165888 $$

Since δ ≠ 0, the equation does not correspond to a degenerate case.

3. Existence analysis.

To check if the equation corresponds to a real or imaginary ellipse, we use the trace invariant:

$$ A + C = 9 + 16 = 25 $$

Since the sign of A + C is different from the sign of δ, the equation corresponds to a real ellipse.

Graph in the Cartesian plane of an ellipse-type conic obtained from the analysis of its general quadratic equation. — Graph of the ellipse-type conic

Find the conic associated with the following general quadratic equation: $ y^2-8x-4y+28=0 $

Equation analysis

Type: Non-degenerate parabola.

Orientation: Horizontal (principal axis parallel to the x-axis).

Standard form: $ \left(y - 2\right)^2 = 8\left(x - 3\right) $

For more information about the conic, check the parabola calculator.

Step-by-step solution

1. Analysis of the conic type.

The given equation is:

$$ y^2 - 8x - 4y + 28 = 0 $$

The general equation of a conic is:

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

By comparing, we extract the coefficients:

$$ A = 0 \quad B = 0 \quad C = 1 \\[0.5em] D = -8 \quad E = -4 \quad F = 28 $$

We use the discriminant to analyze the conic type:

$$ \Delta = B^2 - 4AC \\[0.5em] \Delta = 0 - 4(0)(1) = 0 $$

Since Δ = 0, the equation corresponds to a general parabola.

2. Determinant analysis.

To check if the equation corresponds to a degenerate case, we calculate the following determinant:

$$ \delta = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix} \\[0.5em] \delta = 8ACF - 2AE^2 - 2B^2F + 2BDE - 2CD^2 \\[0.5em] \delta = 8(0)(1)(28) - 2(0)(-4)^2 - 2(0)^2(28) + 2(0)(-8)(-4) - 2(1)(-8)^2 \\[0.5em] \delta = -128 $$

Since δ ≠ 0, the equation does not correspond to a degenerate case.

Graph in the Cartesian plane of a parabola-type conic obtained from the analysis of its general quadratic equation. — Graph of the conic section

Identify which conic curve describes the expression $ y^2-x^2-4x+6y-6=0. $

Equation analysis

Conic section: Non-degenerate hyperbola.

Orientation: Vertical (principal axis parallel to the y-axis).

Standard form: $ \dfrac{\left(y + 3\right)^2}{11} - \dfrac{\left(x + 2\right)^2}{11} = 1 $

For more information about the conic, check the hyperbola calculator.

Step-by-step solution

1. Analysis of the conic type.

The given equation is:

$$ -x^2 + y^2 - 4x + 6y - 6 = 0 $$

The general equation of a conic is:

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

By comparing, we extract the coefficients:

$$ A = -1 \quad B = 0 \quad C = 1 \\[0.5em] D = -4 \quad E = 6 \quad F = -6 $$

We use the discriminant to analyze the conic type:

$$ \Delta = B^2 - 4AC \\[0.5em] \Delta = 0 - 4(-1)(1) = 4 $$

Since Δ > 0, the equation corresponds to a general hyperbola.

2. Determinant analysis.

To check if the equation corresponds to a degenerate case, we calculate the following determinant:

$$ \delta = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix} \\[0.5em] \delta = 8ACF - 2AE^2 - 2B^2F + 2BDE - 2CD^2 \\[0.5em] \delta = 8(-1)(1)(-6) - 2(-1)(6)^2 - 2(0)^2(-6) + 2(0)(-4)(6) - 2(1)(-4)^2 \\[0.5em] \delta = 88 $$

Since δ ≠ 0, the equation does not correspond to a degenerate case.

Graph in the Cartesian plane of a hyperbola-type conic obtained from the analysis of its general equation. — Graph of the hyperbola

Classify the conic whose general equation is $ 5x^2+4xy+2y^2-24=0. $

Equation analysis

Classification: Non-degenerate ellipse.

Orientation: Rotated with respect to the Cartesian axes.

Step-by-step solution

1. Analysis of the conic type.

The given equation is:

$$ 5x^2 + 4xy + 2y^2 - 24 = 0 $$

The general equation of a conic is:

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

By comparing, we extract the coefficients:

$$ A = 5 \quad B = 4 \quad C = 2 \\[0.5em] D = 0 \quad E = 0 \quad F = -24 $$

Since B ≠ 0, the principal axis of the conic has a rotation with respect to the Cartesian axes.

We use the discriminant to analyze the conic type:

$$ \Delta = B^2 - 4AC \\[0.5em] \Delta = (4)^2 - 4(5)(2) = -24 $$

Since Δ < 0, the equation corresponds to a general ellipse.

2. Determinant analysis.

To check if the equation corresponds to a degenerate case, we calculate the following determinant:

$$ \delta = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix} \\[0.5em] \delta = 8ACF - 2AE^2 - 2B^2F + 2BDE - 2CD^2 \\[0.5em] \delta = 8(5)(2)(-24) - 2(5)(0)^2 - 2(4)^2(-24) + 2(4)(0)(0) - 2(2)(0)^2 \\[0.5em] \delta = -1152 $$

Since δ ≠ 0, the equation does not correspond to a degenerate case.

3. Existence analysis.

To check if the equation corresponds to a real or imaginary ellipse, we use the trace invariant:

$$ A + C = 5 + 2 = 7 $$

Since the sign of A + C is different from the sign of δ, the equation corresponds to a real ellipse.

Graph in the Cartesian plane of an ellipse-type conic rotated with respect to the Cartesian axes, obtained from the invariant analysis of its general quadratic equation. — Graph of the rotated ellipse

Given the quadratic equation $ xy=4 $, determine what type of conic it is.

Equation analysis

Classification: Non-degenerate hyperbola.

Orientation: Rotated with respect to the Cartesian axes.

Step-by-step solution

1. Analysis of the conic type.

The given equation is:

$$ xy - 4 = 0 $$

The general equation of a conic is:

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

By comparing, we extract the coefficients:

$$ A = 0 \quad B = 1 \quad C = 0 \\[0.5em] D = 0 \quad E = 0 \quad F = -4 $$

Since B ≠ 0, the principal axis of the hyperbola has a rotation with respect to the Cartesian axes.

We use the discriminant to analyze the conic type:

$$ \Delta = B^2 - 4AC \\[0.5em] \Delta = (1)^2 - 4(0)(0) = 1 $$

Since Δ > 0, the equation corresponds to a general hyperbola.

2. Determinant analysis.

To check if the equation corresponds to a degenerate case, we calculate the following determinant:

$$ \delta = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix} \\[0.5em] \delta = 8ACF - 2AE^2 - 2B^2F + 2BDE - 2CD^2 \\[0.5em] \delta = 8(0)(0)(-4) - 2(0)(0)^2 - 2(1)^2(-4) + 2(1)(0)(0) - 2(0)(0)^2 \\[0.5em] \delta = 8 $$

Since δ ≠ 0, the equation does not correspond to a degenerate case.

Graph in the Cartesian plane of a hyperbola-type conic rotated with respect to the Cartesian axes, obtained from the analysis of its general quadratic equation and invariants. — Graph of the rotated hyperbola

Analyze the equation $ x^2-2xy+y^2-8x=0 $ and find what type of conic it represents.

Equation analysis

Classification: Non-degenerate parabola.

Orientation: Rotated with respect to the Cartesian axes.

Step-by-step solution

1. Analysis of the conic type.

The given equation is:

$$ x^2 - 2xy + y^2 - 8x = 0 $$

The general equation of a conic is:

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

By comparing, we extract the coefficients:

$$ A = 1 \quad B = -2 \quad C = 1 \\[0.5em] D = -8 \quad E = 0 \quad F = 0 $$

Since B ≠ 0, the principal axis of the parabola has a rotation with respect to the Cartesian axes.

We use the discriminant to analyze the conic type:

$$ \Delta = B^2 - 4AC \\[0.5em] \Delta = (-2)^2 - 4(1)(1) = 0 $$

Since Δ = 0, the equation corresponds to a general parabola.

2. Determinant analysis.

To check if the equation corresponds to a degenerate case, we calculate the following determinant:

$$ \delta = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix} \\[0.5em] \delta = 8ACF - 2AE^2 - 2B^2F + 2BDE - 2CD^2 \\[0.5em] \delta = 8(1)(1)(0) - 2(1)(0)^2 - 2(-2)^2(0) + 2(-2)(-8)(0) - 2(1)(-8)^2 \\[0.5em] \delta = -128 $$

Since δ ≠ 0, the equation does not correspond to a degenerate case.

Graph in the Cartesian plane of a parabola-type conic rotated with respect to the Cartesian axes, obtained from the analysis of its general quadratic equation and invariants. — Graph of the rotated parabola

Determine the conic corresponding to the general equation $ x^2+y^2=0 $ and whether it is a degenerate case or not.

Equation analysis

Classification: Degenerate ellipse (a real point).

Step-by-step solution

1. Analysis of the conic type.

The given equation is:

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

The general equation of a conic is:

$$ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$

By comparing, we extract the coefficients:

$$ A = 1 \quad B = 0 \quad C = 1 \\[0.5em] D = 0 \quad E = 0 \quad F = 0 $$

We use the discriminant to analyze the conic type:

$$ \Delta = B^2 - 4AC \\[0.5em] \Delta = 0 - 4(1)(1) = -4 $$

Since Δ < 0, the equation corresponds to a general ellipse.

2. Determinant analysis.

To check if the equation corresponds to a degenerate case, we calculate the following determinant:

$$ \delta = \begin{vmatrix} 2A & B & D \\ B & 2C & E \\ D & E & 2F \end{vmatrix} \\[0.5em] \delta = 8ACF - 2AE^2 - 2B^2F + 2BDE - 2CD^2 \\[0.5em] \delta = 8(1)(1)(0) - 2(1)(0)^2 - 2(0)^2(0) + 2(0)(0)(0) - 2(1)(0)^2 \\[0.5em] \delta = 0 $$

Since δ = 0, the equation corresponds to a degenerate case. In this case, the ellipse is just a point.

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