an ellipse
a circle
a straight line
a parabola
C.
a straight line
As given ⇒ distance of z from origin and point (0,1/3) is same hence z lies on the bisector of the line joining points (0, 0) and (0, 1/3). Hence z lies on a straight line.
A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre of the circle is
an ellipse
a circle
a hyperbola
a parabola
D.
a parabola
A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is 1/2. Then the length of the semi−major axis is
8/3
2/3
5/3
4/3
A.
8/3
Major axis is along x-axis.
A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at
(0, 2)
(1, 0)
(0,1)
(2,0)
B.
(1, 0)
vertex (0,1)
A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is
(x-p)^{2} = 4qy
(x-q)^{2} = 4py
(y-p)^{2} = 4qx
(y-p)^{2} = 4px
A.
(x-p)^{2} = 4qy
In a circle, AB is a diameter where the co-ordinate of A is (p, q) and let the co-ordinate of B is (x_{1} , y_{1} ).
Equation of circle in diameter form is (x - p)(x - x_{1} ) + (y - q)(y - y_{1} ) = 0
x^{2} - (p + x_{1} )x + px_{1} + y^{2} - (y_{1} + q)y + qy_{1} = 0
x^{2} - (p + x_{1} )x + y^{2} - (y_{1} + q)y + px_{1} + qy_{1} = 0
Since this circle touches X-axis
∴ y = 0
⇒ x^{2} - (p + x1 )x + px_{1} + qy_{1} = 0 Also the discriminant of above equation will be equal to zero because circle touches X-axis.
∴ (p + x_{1} )^{2} = 4(px_{1} + qy_{1}) p^{2} + x^{2}_{1} + 2px_{1}
= 4px_{1} + 4qy_{1} x^{2}_{1} - 2px_{1} + p^{2} = 4qy1
Therefore the locus of point B is, (x - p)^{2} = 4qy