Directrix of Hyperbola
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- length of transverse axis =√3 unit
- length of latus rectum =3√3 unit
- length of transverse axis =2√3 unit
- length of latus rectum =6√3 unit
- (x+3)216−(y+2)29=1
- (x+3)29−(y+2)216=−1
- (x+3)27−(y+2)219=−1
- (x+3)29−(y+2)219=1
25(x+2y−3)2−35(2x−y+4)2=1
- 25(2x−y+4)2−35(x+2y−3)2=1
- 2(2x−y+4)2−3(x+2y−3)2=1
- 2(x+2y−3)2−3(2x−y+4)2=1
An ellipse has as semi-minor axis and its foci and the angle is a right angle. Then, the eccentricity of the ellipse is
A hyperbola, having the transverse axis of length , is confocal with the ellipse then, its equation is
- 1
- 2
- 3
- 4
- √3:√2
- 1:2
- 2:1
- 1:√2
- 2a
- 2b
- 2(a−b)
- 2(a+b)
- x+y−3=0
- x+y−5=0
- x+y−1=0
- x+y−7=0
The combined equation of the asymptotes of the hyperbola x2a2−y2b2=1is x2a2−y2b2=−1..
True
False
A sphere
An empty set
A degenerate set
A pair of planes
Show that the line xa+yb=1 touches the curve y = b.e−xa at the point where the curve intersects the axis of Y.
If a chord joining P(a Sec θ, a tan θ), Q(a Sec α, a tan α) on the hyperbola x2−y2=a2 is the normal at P, then T an α
- x2cosec2θ−y2sec2θ=1
- x2sec2θ−y2cosec2θ=1
- x2cos2θ−y2sin2θ=1
- x2sin2θ−y2cos2θ=1
- an ellipse with equation x2b2+y2a2=1
- a hyperbola with equation x2a2−y2b2=1
- a hyperbola with equation x2b2−y2a2=1
- an ellipse with equation x2a2+y2b2=1
- None
- √3
- Option c equation is not clear
- Option a equation is not clear
- Option b equation is not clear
- (8, 12)
- (245, 10)
- (203, 12)
- (8, 10)
- 4√53
- 2√53
- 2√6
- √30
- x2cosec2θ−y2sec2θ=1
- x2sec2θ−y2cosec2θ=1
- x2sin2θ−y2cos2θ=1
- x2cos2θ−y2sin2θ=1
How do you write the standard form of a parabola with the vertex and focus?
In a hyperbola e = 94 and the distance between the directrices is 3. Then the length of Transverse axis is
272
278
274
174
In a hyperbola the distance between the foci is 2 and the distance between directrices is 1 Its eccentricity is
3/2
5/2