Conjugate Hyperbola
Trending Questions
Q. The foci of the ellipse x216+y2b2=1 and the hyperbola x2144+y281=125 coincide. Then the value of b2 is
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Q.
The eccentricity of the hyperbola is
Q. If the angle between the asymptotes of hyperbola x2a2−y2b2=1 is π3. Then the eccentricity of conjugate hyperbola is
Q. If a hyperbola has one focus at the origin and its eccentricity is √2 . One of the directrices is x+y+1=0. Then the equations to its asymptotes are
Q. The eccentricity of the conjugate hyperbola of the hyperbola x2−3y2=1 is
Q.
If the eccentricity of a hyperbola is , then the eccentricity of its conjugate hyperbola is
Q.
If is the eccentricity of the ellipse - and is the eccentricity of the hyperbola , then
Q. If S1 and S2 are the foci of the hyperbola whose transverse axis length is 4 units and conjugate axis length is 6 units, S3 and S4 are the foci of the conjugate hyperbola, then the area of the quadrilateral S1S3S2S4 is sq. units
Q.
The equation of a hyperbola , conjugate to the hyperbola x2+3xy+2y2+2x+3y=0 is
Q.
The foci of the ellipse and the hyperbola coincide. Then the value of is
Q. The equation of the conjugate hyperbola of the hyperbola x2−2y2−2√5x−4√2y−3=0 is
- x2−2y2−2√5x−4√2y+5=0
- x2−2y2−2√5x−4√2y+2=0
- x2−2y2−2√5x−4√2y+3=0
- x2−2y2−2√5x−4√2y−2=0
Q. ntFind the scalar and vector components of (3i+j+3k) in the direction of (i-j-k).n
Q. The eccentricity of the conjugate hyperbola of x216−y29=1 is .
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Q. The equation of the hyperbola whose asymptotes are 3x+4y–2=0, 2x+y+1=0 and which passes
through the point (1, 1) is
through the point (1, 1) is
Q. Let the eccentricity of the hyperbola x2a2−y2b2=1 be reciprocal to that of the ellipse x2+4y2=4. If the hyperbola passes through a focus of the ellipse, then:
- Equation of hyperbola is x23−y22=1
- a focus of hyperbola is (2, 0)
- eccentricity of the hyperbola is 2√3
- Equation of hyperbola is x2−3y2=3
Q. Find magnitude of p if 2i - j + k, i + 3j - 3k and 3i - pj + 5k are co planar
Q. lf f(x)=tanx√1+tan2x, Limx→(π/2)−f(x)=a and Limx→(π/2)+f(x)=b then
- a=b
- a+b=0
- a=1+b
- a+b=2
Q. The domain of the derivative of the function f(x)=⎧⎨⎩tan−1xfor|x|≤112(|x|−1)for|x|>1 is
- R−{1}
- R−{−1, 1}
- R−{−1}
- R−{0}
Q.
A hyperbola x2a2−y2b2=1 is drawn along with its conjugate hyperbola. The foci points of both hyperbolas are connected as shown. Then S1 S3 S2 S4 always forms a
Rhombus
Trapezium
Square
None of the above
Q. The equation of a hyperbola, conjugate to the hyperbola x2+3xy+2y2+2x+3y+1=0, is
- x2+3xy+2y2+2x+3y+1=0
- x2+3xy+2y2+2x+3y+2=0
- x2+3xy+2y2+2x+3y+3=0
- x2+3xy+2y2+2x+3y+4=0
Q. For hyperbola x216−y212=0 and circle x2+y2=25, find concyclic points.
- x=±10√7, y=±5√3√7
- x=±7√10, y=±5√3√7
- x=±10√7, y=±3√5√7
- x=±5√7, y=±10√3√7
Q.
What is the conjugate hyperbola of the hyperbola
x2a2−y2b2=1
x2a2+y2b2=1
x2a2−y2b2=1
x2a2−y2b2=−1
−x2a2−y2b2=1
Q. Consider the parabola x=ay−by2 (where b≠0) If the exhaustive set of values of a for which there exist α, βϵR−{0} such that both the point (α, β) and (β, α) lies on the given parabola is (−∞, p)∪(q, ∞) then p2+q24 is
Q. If a⋅3tanx+a⋅3−tanx−2=0 has real solution, x≠π2, 0≤x≤π, then the set of possible values of the parameter a are
- [−1, 1]
- [−1, 0)
- (0, 1]
- (0, +∞)
Q. The equation of a hyperbola, conjugate to the hyperbola x2+3xy+2y2+2x+3y+1=0, is
Q. If e and e' be the eccentricities of a hyperbola and its conjugate then
1e2+1(e′)2=1.
1e2+1(e′)2=1.
- True
- False
Q. The eccentricity of the conjugate hyperbola of x216−y29=1 is .
- 54
- 2516
- 53
- 259