Answer to Question #91818 in Analytic Geometry for Ra

Question

Which of the following statements are true and which are false? Please give reasons for the answers.
1.) The equation r=acos(θ+α)+bsin(θ+α) represents a circle.
2.) The direction ratios of the line x-5=5-y, z=5 are 1,1,5.
3.) The intersection of a plane and a cone can be a pair of lines.
4.) The angle between the planes x+2y+2z=5 and 2x+2y+3=0 is 60º
5.) The equations 2x²+y²+3z²+4x+4y+18z+34=0, 2x²-y²=4y-4y-4x represent a real conic.

Answer 1

1) True

$$cosine   and  sine  are  periodic, so (\theta+\alpha)  will  represent same figure as  \theta,\\ turned  by  some  angle \\  \\ polar coordinates is: x=rsin\phi,  y=rcos\phi \\ so, in  Cartesian coordinates : \\ r=\sqrt{x^2+y^2},  cos\theta = \frac{y}{r} = \frac{y}{\sqrt{x^2+y^2}},  sin\theta=\frac{x}{r} = \frac{x}{\sqrt{x^2+y^2}} \\ \\ our  equation:r=acos(\theta+\alpha) + bsin(\theta+\alpha) \\ same figure: r=acos(\theta) + bsin(\theta) \\\sqrt{x^2+y^2} = a\frac{y}{\sqrt{x^2+y^2}} + b\frac{x}{\sqrt{x^2+y^2}} \\ \\ x^2+y^2 = ax+by \\x^2-ax+\frac{a^2}{4} + y^2-by+\frac{b^2}{4} - \frac{a^2}{4} - \frac{b^2}{4} = 0\\  \\ (x-\frac{a}{2})^2 + (y-\frac{b}{2})^2 = \frac{a^2+b^2}{4} \\ \\ we got  circle  equation$$

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,

$$cos(X) = \frac{2+4}{\sqrt{1+4+4}\sqrt{4+4}} = \frac{6}{3\sqrt{8}} = \frac{\sqrt{2}\sqrt{2}}{2\sqrt{2}} = \frac{\sqrt{2}}{2}, so angle   is  45 degrees$$

5)False

$$2x²+y²+3z²+4x+4y+18z+34=0 \\  \\(2x^2+4x+2) + (2y^2+4y+2) +(3z^2+18z+27) +3 = 0 \\ \\ (\sqrt{2}x+\sqrt{2})^2+(\sqrt{2}y+\sqrt{2})^2 + (\sqrt{3}z+3\sqrt{3})^2 = -3    \implies no solutions$$

$$2x²-y²=4y-4y-4x \\ \\(2x^2+4x+2) - (y^2) - 2= 0 \\ 2 = (\sqrt{2}x + \sqrt{2})^2 - y^2 \\  \\ 1 = \frac{(\sqrt{2}x + \sqrt{2})^2}{2} - \frac{y^2}{2} \implies its hyperbola$$