document.write( "Question 1147397: find the canonical form and determine the nature of conics
\n" ); document.write( "8x^2-12xy+17y^2+16x-12y+3=0 \n" ); document.write( "
Algebra.Com's Answer #854706 by KMST(5398)\"\" \"About 
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A quadratic equation of the form \"Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0\" , like \"8x%5E2-12xy%2B17y%5E2%2B16x-12y%2B3=0\" , could represent a circle, ellipse, hyperbola, parabola.
\n" ); document.write( "In special cases it could represent a point, a line, a pair of lines, or no point that could exist, depending on the values of the coefficients.
\n" ); document.write( "The value \"B%5E2-4AC\" , called the discriminant, suggests parabola, ellipse, or hyperbola if it is zero, negative, or positive respectively.
\n" ); document.write( "In the case of \"8x%5E2-12xy%2B17y%5E2%2B16x-12y%2B3=0\" , with \"system%28A=8%2CB=-12%2CC=17%29\" ,
\n" ); document.write( "\"B%5E2-4AC=%28-12%29%5E2-4%2A8%2A17=144-544=-400%3C0\" suggest that the equation represents an ellipse.
\n" ); document.write( "If there was no term in \"xy\" , the axes of symmetry of the ellipse would be parallel to the x- an y-axes.
\n" ); document.write( "If there is term in \"xy\" , one of the axes of symmetry will be at a positive angle \"alpha\" to the positive x-axis such that \"0%5Eo%3Calpha%3C90%5Eo\" .
\n" ); document.write( "Rotating the coordinate axes counterclockwise any angle \"alpha\" such that \"0%5Eo%3Calpha%3C90%5Eo\" ,
\n" ); document.write( "a point \"P%28x%2Cy%29\" would be called the point \"P%28u%2Cv%29\" with coordinates referencing the new axes,
\n" ); document.write( "with \"u=x%2Acos%28alpha%29%2By%2Asin%28alpha%29\" and \"v=-x%2Asin%28alpha%29%2By%2Acos%28alpha%29\" .
\n" ); document.write( "For the reverse conversion, \"highlight%28x=u%2Acos%28alpha%29-v%2Asin%28alpha%29%29\" and \"highlight%28y=u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29\".
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\n" ); document.write( "The points represented by equation \"Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0\" would be represented in the new set of axes
\n" ); document.write( "by a new equation in \"u\" and \"v\" , with new coefficients. \"Ju%5E2%2BKuv%2BLv%5E2%2BMu%2BNv%2BF=0\" .
\n" ); document.write( "Substituting \"highlight%28u%2Acos%28alpha%29-v%2Asin%28alpha%29%29\" for\"x\" and \"highlight%28u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29\" for \"y\" into the original equation,
\n" ); document.write( "we would get the equation \"Ju%5E2%2BKuv%2BLv%5E2%2BMu%2BNv%2BF=0\" .
\n" ); document.write( "Expressions to calculate the new coefficients can be found, including we must find a value for \"alpha\" that makes \"K=0\"
\n" ); document.write( "\"B%28cos%5E2%28alpha%29-sin%5E2%28alpha%29%29%2B2%28C-A%29sin%28alpha%29cos%28alpha%29=0\" --> \"B%28cos%5E2%28alpha%29-sin%5E2%28alpha%29%29=2%28A-C%29sin%28alpha%29cos%28alpha%29\" --> \"B%2F%28A-C%29\"\"%22=%22\"\"2sin%28alpha%29cos%28alpha%29%2F%28cos%5E2%28alpha%29-sin%5E2%28alpha%29%29\"
\n" ); document.write( "That can be converted into \"highlight%28B%2F%28A-C%29=sin%282alpha%29%2Fcos%282alpha%29=+tan%282alpha%29%29\" by using the trigonometric identities \"cos%282alpha%29=+cos%5E2%28alpha%29-sin%5E2%28alpha%29\" and \"sin%282alpha%29=+sin%28alpha%29cos%28alpha%29\"
\n" ); document.write( "From there the values for \"2alpha\" \"alpha\" and its trigonometric functions can be found.
\n" ); document.write( "With those values the other coefficients can be calculated as
\n" ); document.write( "\"J=A%2Acos%5E2%28alpha%29+%2BB%2Asin%28alpha%29cos%28alpha%29%2BC%2Asin%5E2%28alpha%29+\" ,
\n" ); document.write( "\"L=+A%2Asin%5E2%28alpha%29+-B%2Asin%28alpha%29cos%28alpha%29%2BC%2Acos%5E2%28alpha%29\" ,
\n" ); document.write( "\"M=+D%2Acos%28alpha%29%2BE%2Asin%28alpha%29\" , and \"N=-D%2Asin%28alpha%29%2BE%2Acos%28alpha%29\" .
\n" ); document.write( "After that, the canonical form for the equation in \"u\" and \"v\" can be found.
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\n" ); document.write( "FINDING \"alpha\" AND ITS TRIGONOMETRIC FUNCTIONS:
\n" ); document.write( "As a quadratic equation of the form \"Ax%5E2%2BBxy%2BCy%5E2%2BDx%2BEy%2BF=0\" , \"8x%5E2-12xy%2B17y%5E2%2B16x-12y%2B3=0\" has \"system%28A=8%2CB=-12%2CC=17%2CD=16%2CE=12%2CF=3%29\"
\n" ); document.write( "\"B%2F%28A-C%29\"\"%22=%22\"\"sin%282alpha%29%2Fcos%282alpha%29\"\"%22=%22\"\"tan%282alpha%29\"\"%22=%22\"\"%28-12%29%2F%288-17%29\"\"%22=%22\"\"%28-12%29%2F%28-9%29\"\"%22=%22\"\"12%2F9\"\"%22=%22\"\"4%2F3%29\"
\n" ); document.write( "A calculator could provide a good approximation of the value of \"2alpha=53.13%5Eo\"(rounded) and that value could be used to get approximated values for \"sin%282alpha%29\" , \"cos%282alpha%29\" , \"alpha\" , \"sin%28alpha%29\" , and \"cos%28alpha%29\" .
\n" ); document.write( "Alternatively, exact values can be found using trigonometric identities.
\n" ); document.write( "Because \"0%5Eo%3Calpha%3C90%5Eo\" we know that \"0%5Eo%3C2alpha%3C180%5Eo\"}, but because \"tan%282alpha%29=4%2F3%3E0\" we conclude \"0%5Eo%3C2alpha%3C90%5Eo\" as well, and al trigonometric functions will be positive.
\n" ); document.write( "From \"sin%282alpha%29%2Fcos%282alpha%29=4%2F3\" <--> \"sin%282alpha%29=%284%2F3%29cos%282alpha%29\" and trigonometric identity \"sin%5E2%282alpha%29%2Bcos%5E2%282alpha%29=1\" we get
\n" ); document.write( "\"%28%284%2F3%29cos%282alpha%29%29%5E2%2Bcos%5E2%282alpha%29=1\" --> \"%2816%2F9%29cos%5E2%282alpha%29%2Bcos%5E2%282alpha%29=1\" --> \"%2816%2F9%2B1%29cos%5E2%282alpha%29=1\" --> \"%2825%2F9%29cos%5E2%282alpha%29=1\" --> \"cos%5E2%282alpha%29=9%2F25\" --> \"cos%282alpha%29=sqrt%289%2F25%29=3%2F5=0.6\" and \"sin%282alpha%29=%284%2F3%29cos%282alpha%29=%284%2F3%29%283%2F5%29=4%2F5=0.8\" .
\n" ); document.write( "From \"cos%282alpha%29=3%2F5\" and the trig identities for half angles we can find \"sin%5E2%28alpha%29\" , \"sin%28alpha%29\" , \"cos%5E2%28alpha%29\" , \"cos%28alpha%29\" , as
\n" ); document.write( " --> \"sin%28alpha%29=sqrt%281%2F5%29=sqrt%285%2F25%29=highlight%28sqrt%285%29%2F5%29\"
\n" ); document.write( " --> \"cos%28alpha%29=sqrt%284%2F5%29=highlight%282sqrt%285%29%2F5%29\"
\n" ); document.write( "We could also get the product \"highlight%28sin%28alpha%29cos%28alpha%29=2%2F5%29\" from \"4%2F5=sin%282alpha%29=2sin%28alpha%29cos%28alpha%29\" --> \"%281%2F2%29%282sin%28alpha%29cos%28alpha%29%29=%281%2F2%29%284%2F5%29=2%2F5\" .
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\n" ); document.write( "CALCULATING \"J\" , \"L\" , \"M\" AND \"N\" :
\n" ); document.write( "Now we can substitute the highlighted values above into the equations to calculate \"J\" , \"L\" , \"M\" , and \"N\" :
\n" ); document.write( "\"J=A%2Acos%5E2%28alpha%29+%2BB%2Asin%28alpha%29cos%28alpha%29%2BC%2Asin%5E2%28alpha%29\"=\"8%2A%284%2F5%29%2B%28-12%29%282%2F5%29%2B17%2A%281%2F5%29\"=\"32%2F5%2B17%2F5-24%2F5=25%2F5=5\" ,
\n" ); document.write( "\"%28rounded%29\" , and
\n" ); document.write( "\"%28rounded%29\" .
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\n" ); document.write( "FINDING THE CANONICAL FORM AND THE NATURE OF THIS CONIC:
\n" ); document.write( "For now, substituting the values found for \"J\" and \"L\" , we can write the equation of the conic in terms of \"u\" and \"v\" as:
\n" ); document.write( "\"5u%5E2%2B20v%5E2%2BMu%2BNv%2B3=0\" ,
\n" ); document.write( "and we can and need to transform it into something of the form \"%28u-h%29%5E2%2Fp%5E2%2B%28v-k%29%5E2%2Fq%5E2=1\" which would be equivalent to
\n" ); document.write( "\"%28%28u-h%29%2Fp%29%5E2%2B%28%28v-k%29+%2Fq%29%5E2=1\" .
\n" ); document.write( "That is the equation of a circle of radius \"1\" , stretched in the u and v directions by factors \"p\" and \"q\" , centered in point (h,k) .
\n" ); document.write( "In other words, that represents an ellipse centered in (h,k) .
\n" ); document.write( "To get that transformation we need to form the squares \"%28u-h%29%5E2=u-2uh%2Bh%5E2\" and \"%2B%28v-k%29%5E2=v-2vk%2Bk%5E2\" from \"5u%5E2%2BMu\" and \"20v%5E2+%2BNv\" respectively .
\n" ); document.write( "\"5u%5E2%2B20v%5E2%2BMu%2BNv%2B3=0\" --> \"5u%5E2%2BMu%2B20v%5E2%2BNv=-3\" --> -->
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\n" ); document.write( "\"M%2F10=4sqrt%285%29%2F10=highlight%280.4sqrt%285%29%29=0.89443\"\"%28rounded%29\" ,
\n" ); document.write( "and \"N%2F40=-8sqrt%285%29%2F40=highlight%28-0.2sqrt%285%29%29=-0.447215\"\"%28rounded%29\" .
\n" ); document.write( "\"%28M%2F10%29%5E2=%280.4sqrt%285%29%29%5E2=0.16%2A5=0.8\"
\n" ); document.write( "\"%28N%2F40%29%5E2=%28-0.2sqrt%285%29%29%5E2=0.04%2A5=0.2\"
\n" ); document.write( "Then, the right hand side term is \"-3%2B5%28M%2F10%29%5E2%2B20%28N%2F40%29%5E2\"=\"-3%2B5%2A%280.4sqrt%285%29%29%5E2%2B20%2A%280.5sqrt%285%29%29%5E2\"=\"-3%2B5%2A0.16%2A5%2B20%2A0.04%2A5=-3%2B4%2B4=5\" , and the equation is
\n" ); document.write( "\"highlight%285%28u%2B0.4sqrt%285%29%29%5E2%2B20%28v%2B%28-0.2sqrt%282%29%29%29%5E2=5%29\"
\n" ); document.write( "From there, we continue:
\n" ); document.write( "\"5%28u%2B0.4sqrt%285%29%29%5E2%2B20%28v-0.2sqrt%285%29%29%5E2=5\" --> \"%28u%2B0.4sqrt%285%29%29%5E2%2B4%28v-0.2sqrt%282%29%29%5E2=1\" -->
\n" ); document.write( "\"%28u-%28-0.4sqrt%285%29%29%29%5E2%2F1%5E2\"\"%22%2B%22\"\"%28v-0.2sqrt%285%29%29%5E2%2F0.5%5E2\"\"%22=%22\"\"1\"
\n" ); document.write( "The center of the ellipse is at \"%22%28%22\"\"-0.4sqrt%285%29\"\"%22%2C%22\"\"0.2sqrt%285%29\"\"%22%29%22\" .\r
\n" ); document.write( "\n" ); document.write( "The numbers squared in the denominators are the semi-major and semi-minor axes of the ellipse.
\n" ); document.write( "The greater one, \"1\" , is the semi-major axis, and the other I the semi-minor axis
\n" ); document.write( "The extreme values for \"u\" happen when \"v-0.2sqrt%285%29=0\" , and then
\n" ); document.write( "\"%28u%2B0.4sqrt%285%29%29%5E2=1\" --> \"u%2B0.4sqrt%285%29=%22+%22%2B-+1\" --> \"u=-0.4sqrt%285%29+%2B-+1\" .
\n" ); document.write( "The y-coordinate of the center and the extremes show the vertices are at a distance of \"1\" from the center .
\n" ); document.write( "Using the equations \"highlight%28x=u%2Acos%28alpha%29-v%2Asin%28alpha%29%29\" and \"highlight%28y=u%2Asin%28alpha%29%2Bv%2Acos%28alpha%29%29\" , we can find the coordinates of the center and vertices of the ellipse.
\n" ); document.write( "For the center, \"x=%28-0.4sqrt%285%29%29%2A%282sqrt%285%29%2F5%29-0.2sqrt%285%29%2A%28sqrt%285%29%2F5%29\"=
\n" ); document.write( "\"-0.8%2A5%2F5-0.04sqrt%2825%29=-0.8-0.04%2A5=-0.8-0.2=-1\"
\n" ); document.write( "\"y=-0.4sqrt%285%29%2A%28sqrt%285%29%2F5%29%2B0.2sqrt%285%29%2A%282sqrt%285%29%2F5%29\"=
\n" ); document.write( "\"-0.4%2A5%2F5%2B0.2%2A2%2A5%2F5=-0.4%2B0.4=0\"\r
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