Question #200467

Which of the following statements are true and which are false? Give reasons for your answer.

i) The equation r = acos θ +b sin θ represents a circle.

ii) If 1,1/2,0 are direction ratios of a line, then the line makes an angle of 90◦ with the x-axis, an angle of 60◦ with the y-axis, and is parallel to the z-axis.

iii) The intersection of a plane and a cone can be a pair of lines. 

iv) If a cone has three mutually perpendicular generators then its reciprocal cone has three mutually perpendicular tangent planes. 

v) The equations 2x 2 +y 2 +3z 2 +4x+4y+18z+34 = 0,2x 2 −y 2 = 4y−4y−4x represent a real conic.



1
Expert's answer
2021-06-02T10:36:52-0400

i)


r=a(xr)+b(yr)r=a(\dfrac{x}{r})+b(\dfrac{y}{r})

r2=ax+byr^2=ax+by

x2+y2=ax+byx^2+y^2=ax+by

(xa2)2+(yb2)2=a24+b24(x-\dfrac{a}{2})^2+(y-\dfrac{b}{2})^2=\dfrac{a^2}{4}+\dfrac{b^2}{4}

This is the equation of circle.

Statement is True.


ii) Let the angles made by the line with positive x axis, y axis and z axis be α,β,γ,\alpha, \beta, \gamma, the direction cosines of the lines are (cosα,cosβ,cosγ)(\cos \alpha, \cos \beta, \cos \gamma)


Given cosα=1,cosβ=1/2,cosγ=0.\cos \alpha=1, \cos \beta= 1/2, \cos \gamma=0.

Then α=0°,β=60°,γ=90°.\alpha=0\degree , \beta=60\degree, \gamma=90\degree.

Statement is False.


iii) Conic section can be parabola, hyperbola. ellipse (circle)

Planes that pass through the vertex of the cone will intersect the cone in a point, a line or a pair of intersecting lines. These are called degenerate conics.

Statement is True.


iv) If a cone has three mutually perpendicular generators then its reciprocal cone has

three mutually perpendicular tangent planes and vice versa.

Statement is True.


v) Transform the equations:


2x2+y2+3z2+4x+4y+18z+34=02x^2 + y^2 + 3z^2 + 4x + 4y + 18z + 34 = 0

2(x2+2x+1)2+(y2+4y+4)42(x^2 +2x+1)-2+ (y^2+4y+4)-4

+3(z2+6z+9)27+34=0+ 3(z^2 +6z+9)-27 + 34 = 0

2(x+1)2+(y+2)2+3(z+3)2=12(x+1)^2+ (y+2)^2+3(z+3)^2=-1


2x2y2=4y4z4x2x^ 2 −y^2 = 4y−4z−4x

4z=2(x2+2x+1)+2+(y2+4y+4)44z=-2(x^2+2x+1)+2+(y^2+4y+4)-4

4z=2(x+1)2+(y2)224z=-2(x+1)^2+(y-2)^2-2

z=12(x+1)2+14(y2)212z=-\dfrac{1}{2}(x+1)^2+\dfrac{1}{4}(y-2)^2-\dfrac{1}{2}

The first of these hasn’t any solution and the second is an equation of the saddle-shaped structure (known as a hyperbolic paraboloid or an anticlastic surface). Hence, the equations do not represent a real conic.

Statement is False.



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