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Consider the conic section -5x² + 6xy + 3y² - 4x - 12y + 9 = 0. Rotate and translate the coordinate axes to write it in standard form and name the type of conic section.

User Chadams
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Final answer:

To express the conic section -5x² + 6xy + 3y² - 4x - 12y + 9 = 0 in standard form, we must eliminate the xy-term through axis rotation and complete the square for translation. The type of the conic section (circle, ellipse, parabola, or hyperbola) is then determined based on the coefficients of the quadratic terms in the transformed equation.

Step-by-step explanation:

To write the conic section -5x² + 6xy + 3y² - 4x - 12y + 9 = 0 in standard form and determine its type, we need to perform a rotation of axes to eliminate the xy-term and then complete the square to find translations that bring the equation to standard form.

The first step is to find the angle θ that will eliminate the xy-term. This angle can be calculated using the formula θ = ½ arctan(±2B/(A-C)) where A is the coefficient of x², B is the coefficient of xy, and C is the coefficient of y². Since we have A = -5, B = 6, and C = 3, substituting these values we get θ = ½ arctan(-6/(-5-3)).

After finding the angle θ, we use the rotation transformation equations x = x'cos(θ) + y'sin(θ) and y = -x'sin(θ) + y'cos(θ) to find the new equation in terms of x' and y'. Once the xy-term is eliminated, we look for the complete squares in terms of x' and y' and adjust the equation accordingly by adding and subtracting terms.

Finally, we translate the axis to remove linear terms, if necessary, to get the conic section in its standard form. Depending on the signs and coefficients of the quadratic terms, and whether or not there is a linear term remaining, the equation will represent a circle, ellipse, parabola, or hyperbola. Each of these has a specific standard form that can be matched to the transformed equation.

User Sverrisson
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