Consider the equation: 23x2−6xy+3x+3y=0 a) Use the discriminant to determine whether the graph of the equation is a parabola, an ellipse, or a hyperbola? b) Use a rotation of axes to eliminate the xy-term. (Write an equation in XY-coordinates. Use a rotation angle that satisfies 0≤φ≤π/2)

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Caren · Accepted Answer

Step 1We have given an equation,23x2−6xy+3x+3y=0the general Cartesian form of a conic section:Ax2+Bxy+Cy2+Dx+Ey+F=0The discriminant is B2−4AC(a) If B2−4AC&amp;lt;0, then the equation represents an ellipse.(b) If B2−4AC=0, then the equation represents a parabola.(c) If B2−4AC&amp;gt;0, then the equation represents a hyperbola.Step 2On comparing the given equation with the general form of conic section, we get,A=23,B=−6,C=0The discriminant is −62−4×23×0=3636&amp;gt;0So the equation is hyperbola.Step 3b) In such a case, the relation between coordinate (x, y) and new coordinates (x&#039;, y&#039;) is given by:x=x′cos⁡θ−y′sin⁡θandy=x′cos⁡θ+y′sin⁡θcot⁡2θ=C−AB=0−23−6cot⁡2θ=C−AB=13cot⁡2θ=C−AB=cot,π3θ=π6sin,π6=12cos,π6=32We shall find the value of x,y by these values.x=x′cos⁡θ−y′sin⁡θandy=x′cos⁡θ+y′sin⁡θx=x′32−y′12andy=x′32+y′12On plugging these values in the equation we get,23x2−6xy+3x+3y=023(x′32−y′12)2−6(x′32−y′12)(x′32+y′12)+3(x′32−y′12)+3(x′32+y′12)=0

Consider the equation:2sqrt(3x)^2-6xy+sqrt(3x)+3y=0 a) Use the discriminant to determine

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