1) True
c o s i n e a n d s i n e a r e p e r i o d i c , s o ( θ + α ) w i l l r e p r e s e n t s a m e f i g u r e a s θ , t u r n e d b y s o m e a n g l e p o l a r c o o r d i n a t e s i s : x = r s i n ϕ , y = r c o s ϕ s o , i n C a r t e s i a n c o o r d i n a t e s : r = x 2 + y 2 , c o s θ = y r = y x 2 + y 2 , s i n θ = x r = x x 2 + y 2 o u r e q u a t i o n : r = a c o s ( θ + α ) + b s i n ( θ + α ) s a m e f i g u r e : r = a c o s ( θ ) + b s i n ( θ ) x 2 + y 2 = a y x 2 + y 2 + b x x 2 + y 2 x 2 + y 2 = a x + b y x 2 − a x + a 2 4 + y 2 − b y + b 2 4 − a 2 4 − b 2 4 = 0 ( x − a 2 ) 2 + ( y − b 2 ) 2 = a 2 + b 2 4 w e g o t c i r c l e e q u a t i o n cosine and sine are periodic, so (\theta+\alpha) will represent same figure as \theta,\\ turned by some angle \\ \\
polar coordinates is: x=rsin\phi, y=rcos\phi \\ so, in Cartesian coordinates : \\ r=\sqrt{x^2+y^2}, cos\theta = \frac{y}{r} = \frac{y}{\sqrt{x^2+y^2}}, sin\theta=\frac{x}{r} = \frac{x}{\sqrt{x^2+y^2}} \\ \\ our equation:r=acos(\theta+\alpha) + bsin(\theta+\alpha) \\ same figure: r=acos(\theta) + bsin(\theta) \\\sqrt{x^2+y^2} = a\frac{y}{\sqrt{x^2+y^2}} + b\frac{x}{\sqrt{x^2+y^2}} \\ \\
x^2+y^2 = ax+by \\x^2-ax+\frac{a^2}{4} + y^2-by+\frac{b^2}{4} - \frac{a^2}{4} - \frac{b^2}{4} = 0\\ \\
(x-\frac{a}{2})^2 + (y-\frac{b}{2})^2 = \frac{a^2+b^2}{4} \\ \\ we got circle equation cos in e an d s in e a re p er i o d i c , so ( θ + α ) w i ll re p rese n t s am e f i gu re a s θ , t u r n e d b y so m e an g l e p o l a r coor d ina t es i s : x = rs in ϕ , y = rcos ϕ so , in C a r t es ian coor d ina t es : r = x 2 + y 2 , cos θ = r y = x 2 + y 2 y , s in θ = r x = x 2 + y 2 x o u r e q u a t i o n : r = a cos ( θ + α ) + b s in ( θ + α ) s am e f i gu re : r = a cos ( θ ) + b s in ( θ ) x 2 + y 2 = a x 2 + y 2 y + b x 2 + y 2 x x 2 + y 2 = a x + b y x 2 − a x + 4 a 2 + y 2 − b y + 4 b 2 − 4 a 2 − 4 b 2 = 0 ( x − 2 a ) 2 + ( y − 2 b ) 2 = 4 a 2 + b 2 w e g o t c i rc l e e q u a t i o n
2)False.
x-5 = 5-y, so y = -x + 10 and in xOy plane directions are {1: -1} or\and {-1: 1}, so, {1:1:5} doesnt fit.
3)False, conic section can be parabola, hyperbola. ellipse and line
4)False
our planes: x+2y+2z=5 and 2x+2y+3=0
we have 2 normal vectors: a=(1,2,2) and b=(2,2,0)
the angle X can be found by the formula cos(X) = (a, b) \ (||a|| * ||b||)
so,
c o s ( X ) = 2 + 4 1 + 4 + 4 4 + 4 = 6 3 8 = 2 2 2 2 = 2 2 , s o a n g l e i s 45 d e g r e e s cos(X) = \frac{2+4}{\sqrt{1+4+4}\sqrt{4+4}} = \frac{6}{3\sqrt{8}} = \frac{\sqrt{2}\sqrt{2}}{2\sqrt{2}} = \frac{\sqrt{2}}{2}, so angle is 45 degrees cos ( X ) = 1 + 4 + 4 4 + 4 2 + 4 = 3 8 6 = 2 2 2 2 = 2 2 , so an g l e i s 45 d e g rees
5)False
2 x 2 + y 2 + 3 z 2 + 4 x + 4 y + 18 z + 34 = 0 ( 2 x 2 + 4 x + 2 ) + ( 2 y 2 + 4 y + 2 ) + ( 3 z 2 + 18 z + 27 ) + 3 = 0 ( 2 x + 2 ) 2 + ( 2 y + 2 ) 2 + ( 3 z + 3 3 ) 2 = − 3 ⟹ n o s o l u t i o n s 2x²+y²+3z²+4x+4y+18z+34=0 \\ \\(2x^2+4x+2) + (2y^2+4y+2) +(3z^2+18z+27) +3 = 0 \\ \\
(\sqrt{2}x+\sqrt{2})^2+(\sqrt{2}y+\sqrt{2})^2 + (\sqrt{3}z+3\sqrt{3})^2 = -3 \implies no solutions 2 x 2 + y 2 + 3 z 2 + 4 x + 4 y + 18 z + 34 = 0 ( 2 x 2 + 4 x + 2 ) + ( 2 y 2 + 4 y + 2 ) + ( 3 z 2 + 18 z + 27 ) + 3 = 0 ( 2 x + 2 ) 2 + ( 2 y + 2 ) 2 + ( 3 z + 3 3 ) 2 = − 3 ⟹ n o so l u t i o n s
2 x 2 − y 2 = 4 y − 4 y − 4 x ( 2 x 2 + 4 x + 2 ) − ( y 2 ) − 2 = 0 2 = ( 2 x + 2 ) 2 − y 2 1 = ( 2 x + 2 ) 2 2 − y 2 2 ⟹ i t s h y p e r b o l a 2x²-y²=4y-4y-4x \\ \\(2x^2+4x+2) - (y^2) - 2= 0 \\ 2 = (\sqrt{2}x + \sqrt{2})^2 - y^2 \\ \\ 1 = \frac{(\sqrt{2}x + \sqrt{2})^2}{2} - \frac{y^2}{2} \implies its hyperbola 2 x 2 − y 2 = 4 y − 4 y − 4 x ( 2 x 2 + 4 x + 2 ) − ( y 2 ) − 2 = 0 2 = ( 2 x + 2 ) 2 − y 2 1 = 2 ( 2 x + 2 ) 2 − 2 y 2 ⟹ i t s h y p er b o l a
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