General Equation of Hyperbola
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Q. 30. the equation of the hyperbola whose asymptotes are the straight lines 3x-4y+7=0 and 4x+3y+1=0 which passes through (0, 0)
Q. If the length of latus rectum of a hyperbola x2k−y225=−1 is 225 units, then its e (eccentricity) is
- 73
- 75
- 65
- 72
Q.
Find the equation for the ellipse that satisfies the given conditions,
b=3, c=4, centre at origin; foci on the x - axis.
Q. A(−2, 0) and B(2, 0) are the two fixed points and P is a point such that PA−PB=2. Let S be the circle x2+y2=r2, then match the following.
Column IColumn IIa. If r=2, then the number of points P satisfying p. 2PA−PB=2 and lying on x2+y2=r2 is b. If r=1, then the number of points P satisfying q. 4PA−PB=2 and lying on x2+y2=r2 is c. For r=2 the number of common tangents is r. 0d. For r=12 the number of common tangents is s. 1
Column IColumn IIa. If r=2, then the number of points P satisfying p. 2PA−PB=2 and lying on x2+y2=r2 is b. If r=1, then the number of points P satisfying q. 4PA−PB=2 and lying on x2+y2=r2 is c. For r=2 the number of common tangents is r. 0d. For r=12 the number of common tangents is s. 1
- a−q; b−p; c−r; d−p
- a−s; b−q; c−r; d−p
- a−p; b−s; c−r; d−p
- a−q; b−p; c−s; d−s
Q. If a variable line has its intercepts on the co-ordinate axes e, e′, where e2, e′2 are the eccentricities of a hyperbola and its conjugate hyperbola, then the line always touches the circle x2+y2=r2, where r=
Q.
Equation of the hyperbola with focus (-3, 4) directrix 3x-4y+5=0 and e = 52 is
5x2−24xy+12y2+6x−8y−75=0
5x2−24xy+12y2−8x−6y−25=0
5x2−24xy+12y2−12x+8y−55=0
5x2−24xy+12y2−7x−12y−65=0
Q. If the chord xcosα+ysinα=p of the hyperbola x216−y218=1 subtends a right angle at the centre, and the diameter of the circle, concentric with the hyperbola to which the given chord is a tangent is d units, then the value of d4 is
Q. If the length of latus rectum of a hyperbola whose eccentricity is 3√5, centre (4, 3) and axis is parallel to coordinate axis is 8 units, then the equation(s) of the hyperbola is/are
- (y−3)225−(x−4)220=1
- (x−4)225−(y−3)220=1
- (x−4)225−(y−3)215=1
- (2x+y−11)225−(x−2y+2)215=1