Definition of Ellipse
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Let a vector be obtained by rotating the vector by an angle about the origin in counter clockwise direction in the first quadrant. Then the area of triangle having vertices and is equal to:
- 32
- √32
- 23
- √23
A hyperbola having the transverse axis of length has the same foci as that of the ellipse , then this hyperbola does not pass through which of the following points -
- the eccentricity is 12
- the length of latus-rectum is 32
- one focus is at (3√3, 0)
- one directrix is x=−2√3
If three points A, B&C have position vectors (1, x, 3), (3, 4, 7) and (y, -2, -5) respectively.If they are collinear find the values of x&y.
- 23
- 19
- 12
- 13
- 1−√71+√7
- √5−1√5+1
- 1−√51+√5
- √7−1√7+1
- 7x2+15y2=247
- 49x2+225y2=105
- 7x2+15y2=105
- 7x2+15y2=147
- √74
- 25
- √35
- √53
- 3x−5y+25=0
- 3x+5y+25=0
- 3x+5y=25
- 3x−5y=25
If the distance between the foci of an ellipse is and the distance between its directrices is , then the length of its latus rectum is
Why is the eccentricity of an ellipse between and ?
(correct answer + 2, wrong answer - 0.50)
- [12, ∞)
- (−∞, −12]∪[12, ∞)
- (−∞, −6]∪[6, ∞)
- [6, ∞)
C1 and C2 are circles of unit radius with centres at (0, 0) and (1, 0) respectively. C3 is a circle of unit radius passes through the centres of the circles C1 and C2 and have its centre above x-axis. Equation of the common tangent to C1 and C3 which does not pass through C2 is
- For the ellipse, the eccentricity is 1√2
- For the ellipse, the eccentricity is 12
- For the ellipse, the length of the latus rectum is 1 unit
- For the ellipse, the length of the latus rectum is 12
eccentricity 34 and directrix 2x−y+3=0 is
- 44x2+36xy+71y2−588x+374y+959=0
- 44x2+36xy+71y2−374x−528y+756=0
- 44x2+36xy+71y2−135x−47y+859=0
- 44x2+36xy+71y2−125x−274y+659=0
Let A(a cos θ, b sin θ) is a variable point, S=(ap, 0) and S′ ≡ (−ap, 0) are two fixed points where p=√a2−b2a2. If the locus of Incentre of triangle ASS' is a conic then the eccentricity of the conic in terms of p is
- √p1+p
- √1−p1+p
- p2(1+p)
- √2p1+p
- x264+y232=1
- x28+y24=1
- x264+y216=1
- x232+y232=1
- 2
- 5
- 10
- 20