Answer to Question #104276 in Analytic Geometry for Gayatri Yadav

Question

Which of the following statements are true, and which are false? Justify your answer
with a short proof or a counter-example.
i) The set of all the points (x, y,z) satisfying the equation x−z = z−y represents a
line.
ii) Any two conics can intersect at at most two points.
iii) The circle (x−1)^2+y^2 = 1,z = 0 lies inside the sphere centred at the origin, and
having radius 2√2.
iv) If a curve C is symmetric about both the coordinate axes, then C is symmetric about
the origin.
v) Every planar section of an ellipsoid is an ellipse.
vi) If α,β, γ are direction ratios of a line, then so are α^2,β^2, γ^2.
vii) The projection of the line segment joining (1,2,−1) and (4,2,−1) on the line
x = y = z is 0.
viii) The polar equation r = θ represents a conic.
ix) If P is a point on an ellipse with a focus F, then PF is always greater than PM,where M is the foot of the perpendicular drawn from P to a directrix of the ellipse.
x) x^2 +y^2 +z^2 = xyz is the equation of a cone.

Answer 1

i)

$x-z=z-y=>x+y-2z=0$

This is an equation of the plane instead of the line. Hence, the initial statement of the question is false.

False.

ii) It is known from algebra that the simultaneous solution set of two equations of the second degree consists of four points. Therefore, two conics will always intersect at four points. These points may all be real and distinct, two real and two imaginary or all imaginary. Two or more points may also coincide.

False.

iiI)

$\begin{cases} (x-1)^2+y^2=1 & \\ z=0 \end{cases}=>\begin{cases} x^2-2x+1+y^2=1 & \\ z=0 \end{cases}$

$=>\begin{cases} x^2+y^2=2x & \\ z=0 \end{cases}=>x^2+y^2+z^2=2x$

We know that $(x-1)^2+y^2=1.$ Then $0\leq(x-1)^2\leq1=>0\leq x\leq2$

We have that $x^2+y^2+z^2=2x\leq2\cdot2=4<8=(2\sqrt{2})^2$

Hence the circle $(x-1)^2+y^2=1, z=0$ lies inside the sphere centred at the origin, and

having radius $2\sqrt{2}.$

True

iv)

If a curve C is symmetric about the coordinate x-axis, then

$(x, y)\to(x, -y), (-x, y)\to(-x, -y), (x, -y)\to(x, y), (-x, -y)\to(-x, y)$

If a curve C is symmetric about the coordinate y-axis, then

$(x, y)\to(-x, y), (-x, y)\to(x, y), (x, -y)\to(-x, -y), (-x, -y)\to(x, -y)$

If a curve C is symmetric about both the coordinate axes

$(x, y)\to(-x, y), (x, y)\to(-x, -y), (x, y)\to(x, -y)$

C is symmetric about the origin. 

True.

v) Every planar cross section is either an ellipse, or is empty, or is reduced to a single point (this explains the name, meaning "ellipse like"). 

Every planar cross section is an ellipse because it is a closed quadratic curve.

True.

vi) As the direction ratio of a line can be positive or negative but $\alpha^2, \beta^2, \gamma^2$ will be always non-negative hence it will not represent the correct direction ratio.

False.

vii) The line segment passing through the points $(1,2,-1)$ and $(4,2,-1)$ is parallel to  $\vec{a}=3\vec{i}.$

Direction ratios of a line $l=\vec{i}+\vec{j}+\vec{k}$

Hence the required projection is

$${\vec{a}\cdot\vec{l} \over |\vec{l}|}={3 \over \sqrt{1^2+1^2+1^2}}\not=0$$

False.

viii)

$r=\theta$

The graph is a spiral that starts from the origin of coordinates (when $\theta=0$) and unravels counterclockwise.

False.

ix) For all points on the ellipse, the ratio between the distance to the focus and the distance to the directrix is a constant. 

If F is a focus of the ellipse, P is a point on the ellipse and M is foot of perpendicular from P on the directrix of the ellipse, then the distances PM and PF form a known ratio, which is

$${PF \over PM}=e$$

where $e$ is eccentricity of the ellipse $0<e<1$

Thus $PF=e\cdot PM=>PF<PM$

False.

x) We study only quadratic cones; i.e. cones having its equation of second degree in x, y and z.

The general equation of second degree equation is,

$$ax^2+by^2+cz^2+2fyz+2gxz+2hxy+2ux+2vy+2wz+d=0$$

Then $x^2+y^2+z^2=xyz$ is not the equation of a cone.

False.