Consider 6x2 4xy 3y2 4square root 5x 8square root 5y 1

Consider 6x^2 + 4xy + 3y^2 - (4square root 5)x + (8square root 5)y + 1 = 0

Find the following:

a) conic type b) rotation angle c) new x\'y\' equation in standard form d) sketch on same set of axes

I have absolutely no idea where to start on this problem. Any help would be much appreciated! please show all the steps if you can.

Solution

(a) If in an equation, no variable is squared, it represents a line. To be a conic section, atleast one of the variable should be squared.

Now, if both the variables are squared, and both squared terms are multiplied by the same number, then its a circle.

If only one of the variables is squared, then it is a parabola.

If Both variables are squared and have the same sign, but they aren\'t multiplied by the same number, then it is an ellipse.

If Both variables are squared, and the squared terms have opposite signs, then it is an hyperbola.

So, in this case the conic type is ellipse.

(b) Now, comparing the equation to standard equation:

A.x^2 + B.xy + C.y^2 + D.x + E.y + F = 0

We find the angle of rotation by the relation:

Cot (2.theta) = (A-C)/B

= (6-3)/4 = 3/4

Hence angle comes out to be = 26.55 Degrees.

(c)

Determine the new equation of the ellipse by calculating the following coefficients:

A=Acos²+Bcossin+Csin²
B=0
C=Asin²Bcossin+Ccos²
D=Dcos+Esin
E=Dsin+Ecos
F=F

The resulting equation: Ax²+Cy²+Dx+Ey+F=0
After writing this equation in the form:

(xx₀)²/a² + (yy₀)²/b² = 1 , We will get:

x₀=D²A

y₀=E²C

a² = 4FAC+CD²+AE²/ 4AC²

b² = 4FAC+CD²+AE²/ 4A²C

The coordinates of the centerpoint are found by rotating back about angle

x0=x₀cos y₀sin
y0=x₀sin + y₀cos