Question

1 points Use the Principal Axes Theorem to perform a rotation of ases to eliminate the xy-term in the quadratic equation 22° +12y – 3y-50 0, identify the resulting rotated conic and give is op the new coordinate system a Ellipse: -9(x"}+967"-50=0 Ellipse, 7(x")+6019-50 - 0 Hyperbola: -7x"}+6("2 - 50 = 0 d. Ellipse;9(x")+909")2-50 - 0 e Hyperbola: 7(x") - 609") - 50 = 0

Answer 1

2x2 + 12xy -3y2- 50 20 equation. let A= matrix of giren be the given quadratic 2 6 associated be the 6 -3 quadratic equation, equation of A is 2-34 6 -3-21 characteristic 22 (2-4) (-3-2) -36 = 0 -6-22 +32 +22-36=0 ०४ x3+x42=0 x2 +7X-6 X -9220 or (x+7) CX-6)=0 2: - -7 and a=6. or so are two eigen values -7,6 Let V2 = be the eigen vector corresponding eigen value -7. then g AV2 = -7 V gives + [:). 3a+2bzo therefore put bot a t be equivalent equation is then -t, the eigen beetor kes real number, take t= 3 then Let V2= [e] be the eigen vector AV2=64₂ F3]. corresponding eigen balue 6. gives to $][] =183) then -ze +3d=0 take d=k .KER equad valent equation is then @=k,

put k=z and get V₂ = 3 construet the orthogonal matrix J22:32 - 13 Now charae the coordinate Such that х = (3) = [5 3 — (1 1 23 2 (5) het, az €т г 3х+2а Гі» put x - - 2x'+28 and y het ne ६ given equation л 13 2х2 +12xa -32* -50 = 0 2 (-2x+34*, 42 .-2x+3° зх'+2° 3. ex't y* - 50 - 0 Таз Гіз 15 13 o2 + 2 ( 4х2+ 02.12x'a's +12 (-6х2+9х2-4x''+2+2) - 3 (9х2+ 42 + 12х)- 23. 50 =0 @x2.12x7. 27х2 + 1822 + 772 - - 122 - 650 — -9 x2 + 782 - 456 - 0 - 7х2 + 62 Бо – о this above be the equation of hyperbola. So the correct option is © нарет bolo : -7es++ 4 - 50 =o.