Equation of hyperbola having conjugate axis parallel x axis and ecentricity 2
Answers
Hey mate...
here is your answer
Hyperbola formulae will help us to solve different types of problems on hyperbola in co-ordinate geometry.
1. x2a2x2a2 - y2b2y2b2 = 1, (a > b)
(i) The co-ordinates of the centre are (0, 0).
(ii) The co-ordinates of the vertices are (± a, 0) i.e., (-a, 0) and (a, 0).
(iii) The co-ordinates of the foci are (± ae, 0) i.e., (- ae, 0) and (ae, 0)
(iv) The length of transverse axis = 2a and the length of conjugate axis = 2b.
(v) The transverse axis is along x axis and the equations of transverse axes is y = 0.
(vi) The conjugate axis is along y axis and the equations of conjugate axes is x = 0.
(vii) The equations of the directrices are: x = ± aeae i.e., x = - aeae and x = aeae.
(viii) The eccentricity of the hyperbola is b22 = a22(e22 - 1) or, e = 1+b2a2−−−−−√1+b2a2.
(ix) The length of the latus rectum 2 ∙ b2ab2a = 2a(e22 - 1).
(x) The distance between the two foci = 2ae.
(xi) The distance between two directrices = 2 ∙ aeae.
(xii) Focal distances of a point (x, y) are a ± ex
(xiii) The co-ordinates of the four ends of latera recta are (ae, b2ab2a), (ae, -b2ab2a), (- ae, b2ab2a) and (- ae, -b2ab2a).
(xiv) The equations of latera recta are x = ± ae i.e., x = ae and x = -ae.
2. x2b2x2b2 - y2a2y2a2 = 1, (a > b)
(i) The co-ordinates of the centre are (0, 0).
(ii) The co-ordinates of the vertices are (0, ± a) i.e., (0, -a) and (0, a).
(iii) The co-ordinates of the foci are (0, ± ae) i.e., (0, - ae) and (0, ae)
(iv) The length of transverse axis = 2a and the length of conjugate axis = 2b.
(v) The transverse axis is along Y-axis and the equations of conjugate axes is x = 0.
(vi) The transverse axis is along X-axis and the equations of conjugate axes is y = 0.
(vii) The equations of the directrices are: y = ± aeae i.e., y = - aeae and y = aeae.
(viii) The eccentricity of the hyperbola is b2 = a22(e22 - 1) or, e = 1+b2a2−−−−−√1+b2a2
(ix) The length of the latus rectum 2 ∙ b2ab2a = 2a (e22 - 1).
(x) The distance between the two foci = 2ae.
(xi) The distance between two directrices = 2 ∙ aeae.
(xii) Focal distances of a point (x, y) are a ± ey
(xiii) The co-ordinates of the four ends of latera recta are (b2ab2a, ae), (-b2ab2a, ae), (b2ab2a, -ae) and (-b2ab2a, -ae).
(xiv) The equations of latera recta are y = ± ae i.e., y = ae and y = -ae.
3. (x−α)2a2(x−α)2a2 - (y−β)2b2(y−β)2b2 = 1, (a > b)
(i) The co-ordinates of the centre are (α, β).
(ii) The co-ordinates of the vertices are (α ± a, β) i.e., (α - a, β) and (α + a, β).
(iii) The co-ordinates of the foci are (α ± ae, β) i.e., (α - ae, β) and (α + ae, β)
(iv) The length of transverse axis = 2a and the length of conjugate axis = 2b.
(v) The transverse axis is along parallel to x axis and the equations of transverse axes is y = β.
(vi) The conjugate axis is along parallel to y axis and the equations of conjugate axes is x = α.
(vii) The equations of the directrices are: x = α ± aeae i.e., x = α - aeae and x = α + aeae.
(viii) The eccentricity of the hyperbola is b22 = a22(e22 - 1) or, e = 1+b2a2−−−−−√1+b2a2
(ix) The length of the latus rectum 2 ∙ b2ab2a = 2a (e22 - 1).
(x) The distance between the two foci = 2ae.
(xi) The distance between two directrices = 2 ∙ aeae.
4. (x−α)2b2(x−α)2b2 - (y−β)2a2(y−β)2a2 = 1, (a > b)
(i) The co-ordinates of the centre are (α, β).
(ii) The co-ordinates of the vertices are (α, β ± a) i.e., (α, β - a) and (α, β + a).
(iii) The co-ordinates of the foci are (α, β ± ae) i.e., (α, β - ae) and (α, β + ae).
(iv) The length of transverse axis = 2a and the length of conjugate axis = 2b.
(v) The transverse axis is along parallel to Y-axis and the equations of transverse axes is x = α.
(vi) The conjugate axis is along parallel to X-axis and the equations of conjugate axes is y = β.
(vii) The equations of the directrices are: y = β ± aeae i.e., y = β - aeae and y = β + aeae.
(viii) The eccentricity of the hyperbola is b22 = a22(e22 - 1) or, e = 1+b2a2−−−−−√1+b2a2
(ix) The length of the latus
hope it helps you!!!!
please mark my answers as the brainliest answer.
Answer:
itna lamba answer bhai kahe se copy kr liya h bro