theta=45 degrees to write 2x^2+2xy+5y^2+6x-4y=40 in the x'y'-plane. Then identify the conic.

A figure made up of a surface and a round cone intersecting. A conic section can be a circle, an ellipse, a parabola, or a hyperbola, based on the direction of the plane with regard to the cone.

The angle θ is 45° and the equation is 2x² + 2xy + 5y² + 6x - 4y = 40.

Then x will be

[tex]\rm x = x' \cos \theta - y' \sin \theta\\\\x = x' \cos 45- y' \sin 45\\\\x = \dfrac{x' - y'}{\sqrt2}[/tex]

Then y will be

[tex]\rm y = x' \sin \theta + y' \cos \theta\\\\y= x' \sin 45 + y' \cos 45\\\\y = \dfrac{x' + y'}{\sqrt2}[/tex]

Put these in the equation we have

[tex]\rm 2\left ( \dfrac{x' - y'}{\sqrt2} \right )^2 + 2\left ( \dfrac{x' - y'}{\sqrt2} \right )\left ( \dfrac{x' + y'}{\sqrt2} \right ) + 5 \left ( \dfrac{x' + y'}{\sqrt2} \right )^2+ 6\left ( \dfrac{x' - y'}{\sqrt2} \right ) - 4\left ( \dfrac{x' + y'}{\sqrt2} \right ) = 40[/tex]

On simplifying, we have

[tex]\rm 9x'^2 + 6x'y' + 5y'^2 + 10\sqrt2x' - 10 \sqrt2 y' - 80=0[/tex]

Then the value of a, b, and c will be

a = 2, b = 2, and c = 5

Then we have

[tex]\rm b^2 - 4ac = 2^2 - 4\times 2 \times 5 = 4 - 20 = -16[/tex]

Then b² - 4ac < 0, then the conic is ellipse.

More about the conic section link is given below.

https://brainly.com/question/10311514

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