(a) If a point "M(t,s)" belong to the curve "x^4 + y^4 = 4a^2xy", then "t^4 + s^4 = 4a^2ts". Then for the point "M'(-t,-s)" we have "(-t)^4 + (-s)^4 =t^4+s^4= 4a^2ts=4a^2(-t)(-s)", and thus the point "M'(-t,-s)" is also belong to the curve "x^4 + y^4 = 4a^2xy". Consequently, the curve is symmetric with respect to the origin.
Answer: true
(b) Since the vector "v=(1,-1,0)" is parallel to the line "x = - y,\\ z=0" and "|v|=\\sqrt{2}", we conclude that "\\cos \\alpha =\\frac{1}{\\sqrt{2}}, \\ \\cos \\beta =-\\frac{1}{\\sqrt{2}}, \\ \\cos \\gamma =\\frac{0}{\\sqrt{2}}=0."
Answer: true
(c) The general equation of a hyperbola is
"\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1".
Let us find the section of "2x^2+y^2=2(1-z^2)" by the plane "x+2=0":
"2(-2)^2+y^2=2(1-z^2)"
"8+y^2=2-2z^2"
"y^2+2z^2=-6"
Since the eqution has no real solution, the plane "x+2=0" does not intersect "2x^2+y^2=2(1-z^2)". The equation "y^2+2z^2=-6" is not a hyperbola equation.
Answer: false
(d) The xy-plane intersects the sphere "x^2+y^2+z^2+2x+2y-z=2" in a great circle if and only if the center of this sphere belong to "xy"-plane. Let us rewrite the equation of the sphere in the the following form: "(x+1)^2+(y+1)^2+(z-\\frac{1}{2})^2=2+1+1+\\frac{1}{4}=\\frac{17}{4}." It follows that "M(-1,-1,\\frac{1}{2})" is the center of the sphere. Taking into account that the third coordinate of "M" is not equal to 0, we conclude that the center of the sphere does not belong to the "xy"-plane, and therefore, the sphere "x^2+y^2+z^2+2x+2y-z=2" does not intersect the "xy"-plane in a great circle.
Answer: false
(e) Let consider any line segment "AB" with "|AB|=2." The projection of a line segment "AB" on another line is the line segment "CD" formed by the projections of the end points of the line segment "AB" on this line. If we choose such a line "l" that the segment "AB" is perpendicular to "l", then the points A and B projects on the same point "C=D" of the line "l", and therefore, "|CD|=0".
Answer: false
Comments
Leave a comment