Parametric Equation of Parabola
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If a circle passes through and , then the coordinates of its centre are
The points of intersection of the curves whose parametric equations are x=t2+1, y=2t and x=2s, y=2s Is given by
(2, 2)
(-2, 4)
(1, 2)
(1, - 3)
The line x−2y=0 will be a bisector of the angle between the lines represented by the equation x2−2hxy−2y2=0, if h=
2
−2
12
−12
- (1, 1)
- (2, 2)
- (12, 12)
- (3, 3)
The parametric equations of a parabola are x=t2+1, y=2t+1.The cartesian equation of its directrix is
x+1=0
y=0
none of these
x=0
What is the area bounded by the curve and the line .
sq units
sq units
sq units
None of these
- x2+y2−4x+8y+12=0
- x2+y2−x+4y−12=0
- x2+y2−x4+2y−24=0
- x2+y2−4x+9y+18=0
- x2=y
- y2=x
- y2=2x
- x2=2y
- None of these
If a double ordinate of the parabola y2=4ax be of length 8a, then the angle between the lines joining the vertex of the parabola to the ends of this double ordinate is
- 30∘
- 60∘
- 90∘
- 120∘
- x=4−3t2, y=2−6t
- x=2+3t, y=4+t2
- x=4+3t2, y=2+6t
- x=2+3t2, y=4+6t
- x(y′)2=2yy′−x
- x(y′)2=y
- x(y′)2=x−2yy′
- x(y′)2=2yy′+x
- x=6t , y=2−3t2
- x=1+6t , y=2+3t2
- x=6t , y=3t2
- x=1+6t , y=2−3t2
Find the area of the region enclosed by the parabola x2=y, the line y = x + 2 and the X - axis.
- Vertex is (1, 1)
- Vertex is (−1, −1)
- Length of latus rectum of parabola is 8 units
- Length of latus rectum of parabola is 4 units
- 4x2+y2−4kx=0
- x2+y2−4kx=0
- 2x2+4y2−8kx=0
- 4x2−y2+4kx=0
- y=2
- x+2y=5
- x+y=3
- x=1
- a
- 2a
- 9y2=−6x−8
- 9y2=6x+8
- 9y2=6x−8
- 9y2=−6x+8
Find the locus the mid-point of focal chords of the hyperbola x2a2−y2b2=1?
- 23
- 1
- 16
- 13
- x2+y2−αx−βy=0
- x2−y2+2αx+2βy=0
- αx+βy±√(α2+β2)=0
- x2α2+y2β2=1
- (1, 1)
- (0, 0)
- (−1, 0)
- (0, 1)
- (2t2+2, 2t−3)
- (12t2−2, t−3)
- (12t2−2, t+3)
- (12t2+2, t+3)
- 2(4ab−5c)
- 3(4ab−5c)
- 3(5ab−4c)
- 2(5ab−4c)
Consider the two curves, then
and touch each other at only one point
and touch each other at exactly two points
and intersect (but do not touch ) at exactly two points
and neither intersect nor touch each other