17 π₯ 2 + 18 π₯ π¦ β 7 π¦ 2 β 16 π₯ β 32 π¦ β 18 = 0 17π₯^2 + 18π₯π¦ β 7π¦^2 β 16π₯ β 32π¦ β 18 = 0 17 x 2 + 18 x y β 7 y 2 β 16 x β 32 y β 18 = 0
The rotation angle:
t a n 2 ΞΈ = B A β C tan2\theta=\frac{B}{A-C} t an 2 ΞΈ = A β C B β
where A , B and C are the corresponding coefficients from the equation:
A x 2 + B x y + C y 2 + D x + E y + F = 0 Ax^2+Bxy+Cy^2+Dx+Ey+F=0 A x 2 + B x y + C y 2 + D x + E y + F = 0
t a n 2 ΞΈ = 18 1 + 7 = 3 4 = 2 t a n ΞΈ 1 β t a n 2 ΞΈ tan2\theta=\frac{18}{1+7}=\frac{3}{4}=\frac{2tan\theta}{1-tan^2\theta} t an 2 ΞΈ = 1 + 7 18 β = 4 3 β = 1 β t a n 2 ΞΈ 2 t an ΞΈ β
t a n ΞΈ = 1 3 tan\theta=\frac{1}{3} t an ΞΈ = 3 1 β
The required transformation equations needed to rotate the coordinate axes to eliminate the xy
term in the given equation for the conic section are:
x = x β² c o s ΞΈ β y β² s i n ΞΈ x=x'cos\theta-y'sin\theta x = x β² cos ΞΈ β y β² s in ΞΈ
y = x β² c o s ΞΈ + y β² s i n ΞΈ y=x'cos\theta+y'sin\theta y = x β² cos ΞΈ + y β² s in ΞΈ
c o s ΞΈ = 3 / 10 , s i n ΞΈ = 1 / 10 cos\theta=3/\sqrt{10}, sin\theta=1/\sqrt{10} cos ΞΈ = 3/ 10 β , s in ΞΈ = 1/ 10 β
x = 3 x β² β y β² 10 , y = x β² + 3 y β² 10 x=\frac{3x'-y'}{\sqrt{10}}, y=\frac{x'+3y'}{\sqrt{10}} x = 10 β 3 x β² β y β² β , y = 10 β x β² + 3 y β² β
Substituting these expressions for x and y into the original equation for the conic section and simplifying, we get:
20 x β² 2 β 10 y β² 2 β 80 10 x β² β 80 10 y β² β 18 = 0 20x^{'2}-10y^{'2}-\frac{80}{\sqrt{10}}x'-\frac{80}{\sqrt{10}}y'-18=0 20 x β² 2 β 10 y β² 2 β 10 β 80 β x β² β 10 β 80 β y β² β 18 = 0
Simplifying and completing the squares to obtain an equation for a hyperbola in standard form, we get:
( x β² β 2 / 10 ) 2 1 / 2 β ( y β² + 4 / 10 ) 2 1 = 1 \frac{(x'-2/\sqrt{10})^2}{1/2}-\frac{(y'+4/\sqrt{10})^2}{1}=1 1/2 ( x β² β 2/ 10 β ) 2 β β 1 ( y β² + 4/ 10 β ) 2 β = 1
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