Question

Use the Principal Axes Theorem to perform a rotation of axes to eliminate the xy-term in the quadratic equation 18x2 + 12xy + 13y2 – 48 = 0. Identify the resulting rotated conic and give its equation in the new coordinate system. a Ellipse; 9(x')? +25(v')2 – 48=0 O b. Hyperbola: 10(x')? – 2267')2 - 48 = 0 c. Hyperbola: 9(x")? – 22(y')2 - 48=0 O d. Ellipse: 22(x')> +10(y')2 - 48 = 0 e. Ellipse; 9(x')? +22(")2 – 48=0

Answer 1

We have to eliminate the term xy in the quadratic equation 18x2 + 12 xy + 1342 - 48=0 (11 Let us 9 y sino f rotate the axis of this equation through the angle o. het new coordinate be (x' qy') Then x= x'cos e Y = x'sine + y' coso putting these values in equation (1). we get 18 ( ac' cose - y'sino) +12 (ac'coso - y'sino) (a'sino+y'cose) + 13 (x'sino + y'cose)? -48=0 = 18 cx')?coso + 18 (30) sin?o - 36x'3' sino coso + 12 (x'?cososine - 12 20'y's inap + 12 x'y cosło -52(4')? sino cose + 13 (c' )*sino + 13(y" ? cos?o + 26 x'y' sino coso -48 = 0 (*) equating the coefficient of a'y! to zero -36 sino cose 12 sin?o + 12 cos?o + 26 sino coso - to sino cose + 12 ( cos?o - sino) = 0 1 + 12 COS 20 = -5 sin 20 O U 5 sin 20 = 12005 20 10 11 tanzo 12 5 Thus COS 20 Je Jastat van 11 5 52 +122 5 13 25 +144 | 169 CO320 = a 5 13 and I sin 20 = 12 13

Thus . I +COS20 2 2 2 2 + 72 (30'- 72(31)2 equation (*) can be written as 18 CX')? cosao + 48 (Yu)?sino + 6 (Xu?sin 20 - 6 (40?sin2e + 13(25)2 sine +13(7")?cos? g - 48=0 cos²o f sinzo = 1 - Cos 20 18 (200)2 (1+cos 20) + 1864')? (1-C0520) + 6(x")?sin 20 -GCY J2 sin20 + 1368')? (1- Cos20) + 13(1') {1+60520)_48 = 0 putting sin 20 = 12 f Corvo - 5 in above equation 9 (26")? (1+ *(1-) +6 (29* 13 -6(43% 13 +43(2)*(1-1) + V7*(**) - 48 == +52 (20)2 + 117 (Y!)2 - 624 = 0 286 (31)2 + 117 (40)2-624 =0 2 2 13 1 + → IL 22 (1')2 + 9C%! - 48 = 0 Thus option ces is correct because of y' is just a motation of transformation.