General Equation of Hyperbola
Trending Questions
Q. The equation of the transverse and conjugate axis of the hyperbola 16x2−y2+64x+4y+44=0 are
- x=2, y+2=0
- x=2, y=2
- x+2=0, y=2
- x+2=0, y+2=0
Q. For the hyperbola 9x2−16y2−18x+32y−151=0
- one of the directrix is x=215
- length of latus rectum=92
- foci are (6, 1) and (−4, 1)
- eccentricity is 54
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 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
Q. The equation of the hyperbola whose vertices are (±6, 0) and one of its directrix is x=4 is
x245−y236=1
- x236−y245=1
- x225−y229=1
- y225−x236=1
Q.
The eccentricity of an ellipse whose centre is at the origin is . If one of its directrices is , then the equation of the normal to it at is
Q. The equation of the hyperbola whose foci are at (4, 6), (4, −4) respectively and having eccentricity 2 is
- (y−1)225/8−(x−4)275/4=1
- (y−1)225/4−(x−4)275/4=1
- (y−1)275/4−(x−4)225/4=1
- (y−1)275/4−(x−4)225/8=1
Q.
x2−y2+5x+8y−4=0 represents
Rectangular hyperbola
Ellipse
Hyperbola with centre at (1, 1)
Pair of lines
Q. The equation of the hyperbola whose foci are at (4, 6), (4, −4) respectively and having eccentricity 2 is
- (y−1)225/8−(x−4)275/4=1
- (y−1)225/4−(x−4)275/4=1
- (y−1)275/4−(x−4)225/4=1
- (y−1)275/4−(x−4)225/8=1
Q. The lengths of the transverse axis and the conjugate axis of the hyperbola 9x2−y2=1 are and respectively.
- 2
- 4
- 23
- 43