General Equation of Hyperbola
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The distance between the foci of a hyperbola is 16 and its ecentricity is √2, then equation of the hyperbola is
x2+y2=16
x2−y2=32
x2−y2=16
x2+y2=32
- equation of ellipse is x2+2y2=4
- coordinates of foci of ellipse are (±1, 0)
- equation of ellipse is x2+2y2=2
- coordinates of foci of ellipse are (±√2, 0)
Equation of the ellipse whose axes are the axes of coordinates and which passes through the point and has eccentricity is
The equation of the hyperbola whose foci are (6, 4) and (-4, 4) and eccentricity 2. is
(x−1)225/4−(y−4)275/4=1
(x+1)225/4−(y+4)275/4=1
(x−1)275/4−(y−4)225/4=1
none of these
The area of the triangle formed by the coordinate axes and a tangent to the curve at the point on it is
Find the centre, eccentricity, foci and directrices of the hyperbola(i) 16x2−9y2+32x+36y−164=0(ii) x2−y2+4x=0(iii) x2−3y2−2x=8
- 73
- 75
- 65
- 72
Find the equation of the hyperbola, referred to its principal axes as axes of coordinates, in the following cases:(i) the distance between the foci =16 and eccentricity=√2(ii) conjugate axis is 5 and the distance between foci=13(iii) conjugate axis is 7 and passes through the point (3, -2)
- None of these
- Both A and B
Let and be the eccentricities of the ellipse, and the hyperbola, respectively satisfying . If and are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair is equal to:
x24−y22=1 at the point (x1, y1). Then x21+5y21 is equal to :
- 6
- 10
- 8
- 5
- x2−16xy−11y2−12x+6y+21=0
- 3x2+16x+15y2−4x−14y−1=0
- x2+16x+11y2−12x−6y+21=0
- 3x2+16x+15y2−4x−14y+1=0
- Focus of the parabola is (1, 0)
- The equation of circumcircle of the triangle formed by the tangent and normal at point P and axis of parabola is x2+y2−2x=31
- The length of latus rectum of y2=8√2x and the given parabola is equal
- The area of the quadrilateral formed by the tangents and normals at the extremities of the latus rectum of the given parabola is 64 sq. units
Equation of the hyperbola whose vertices are (±3, 0) and foci at (±5, 0) is
9x2−25y2=81
16x2−9y2=144
9x2−16y2=144
25x2−9y2=225
A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PA-PB=k(k≠0), then the locus of P is
a hyperbola
a branch of the hyperbola
an ellipse
a parabola
- a2+b2=16
- end points of latus rectum will be (±4, ±6)
- ratio of lengths of transverse axis to conjugate axis is 1:√3
- length of latus rectum of the hyperbola =12 unit
- y225−x236=1
x245−y236=1
- x236−y245=1
- x225−y229=1
The equation of the conic with focus at (1, -1) directrix x-y+1=0 and eccentricity√2 is
x2−y2=1
2xy−4x+4y+1=0
xy=1
2xy+4y−4y−1=0
- x+2=0, y+2=0
- x=2, y+2=0
- x=2, y=2
- x+2=0, y=2
In each of the following find the equation of the hyperbola satisfying the given conditions.:
(i)vertices(±2, 0), foci(±3, 0)
(ii)vertices(0, ±, 5), foci(0, ±, 8)
(iii)vertices(0, ±, 3), foci(0, ±, 5)
(iv)vertices(±5, 0), transverseaxis=8
(v)foci(0, ±13), conjugateaxis=24
(vi)foci(±3√5), thelatus−rectum=8
(vii)foci(±, 4, 0), thelatus−rectum=12
(viii)vertices(0, ±6), e=53.
(ix)foci(0, ±√10), passingthrough(2, 3)
(x)foci(0, ±12), latus−rectum=36
The equation of the directrix of a hyperbola is x-y+3 =0.Its ficys us(-1, 1) and eccentricity 3.Find the equation of the hyperbola.
- √3
- √2
- √5
- √6
- a=2
- a=−4
- Equation of circle through the points of intersection of two conic is x2+y2=5
- Equation of circle through the points of intersection of two conic is x2+y2=25
The equation of the hyperbola whose centre is(6, 2) one focus is (4, 2) and of eccenticity 2 is
(x−6)2−2(y−2)2=1
3(x−6)2−(y−2)2=3
(x−6)2−3(y−2)2=1
2(x−6)2−(y−2)2=3
- length of the transverse axis is 2√3
- length of each latus rectum is 32/√3
- eccentricity is √19/3
- equation of a directrix is x=√193
Conjugate axis is 7 and passes through the point (3, −2)
Find the equation of the hyperbola whose
(i) focus is at (5, 2), vertex at (4, 2) and centre at(3, 2)
(ii) focus is at (4, 2), centre at (6, 2) and e=2.