Latus Rectum of Hyperbola
Trending Questions
Q. The point of intersection of two tangents to the hyperbola x2a2−y2b2=1, the product of whose slopes is c2, lies on the curve.
Q.
For some, if the eccentricity of the hyperbola, is times the eccentricity of the ellipse, , then the length of the latus rectum of the ellipse, is:
Q. The eccentricity of the hyperbola −x2a2+y2b2=1 is
Q. Let P be any point on the hyperbola x2a2−y2b2=1 such that the absolute difference of the distances of P from the two foci is 12. If the eccentricity of the hyperbola is 2, then the length of the latus rectum is
- 4√3 unit
- 18 unit
- 2√3 unit
- 36 unit
Q.
The equation of the hyperbola having its eccentricity and the distance between its foci is , is:
Q. Equation of one of the latus rectum of the hyperbola (10x−5)2+(10y−2)2=9(3x+4y−7)2 is
- 30x+40y−23=0
- 30x+40y+23=0
- 40x+30y−23=0
- 40x+30y+23=0
Q. The area of the figure bounded by the parabolas x=−2y2 and x=1−3y2 is
- 43 square units
- 23 square units
- 37 square units
- 67 square units
Q. The equation of the latus rectum of the parabola represented by equation y2+2Ax+2By+C=0 is
Q.
The latus-rectum of the hyperbola
16x2−9y2=144 is
16/3
32/3
8/3
4/3
Q. A is the vertex of the hyperbola x2−2y2−2√5x−4√2y−3=0, B is one of the end points of latus rectum and C is the focus of the hyperbola. If A, B and C lies on same side of conjugate axis, then the area of the triangle ABC is
- 2 sq. units
- √32−1 sq. units
- 1−√23 sq. units
- √32+1 sq. units
Q. If the coordinates of the vertices of a hyperbola are (9, 2) and (1, 2) and the distance between its foci is 10 units. Then
- length of its latus rectum is 9 units
- length of its latus rectum is 92 units
- equation of the hyperbola is (x−5)29- \dfrac{(y-2)^2}{16}=1$
- equation of the hyperbola is (x−5)216- \dfrac{(y-2)^2}{9}=1$
Q. Find the equation of ellipse whose eccentricity is 23, latus rectum is 5 and the centre is (0, 0).
Q. An ellipse intersects the hyperbola 2x2−2y2=1 orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinates axes, then
- the foci of ellipse are (±√2, 0)
- Equation of ellipse is x2+2y2=1
- the foci of ellipse are (±1, 0)
- equation of ellipse are x2+2y2=4
Q. Let P be any point on the hyperbola x2a2−y2b2=1 such that the absolute difference of the distances of P from the two foci is 12. If the eccentricity of the hyperbola is 2, then the length of the latus rectum is
- 4√3 unit
- 18 unit
- 2√3 unit
- 36 unit
Q. The locus of the point of interaction of two tangents of the hyperbola x2a2−y2b2=1 which make an angle α with one another is
Q.
If a hyperbola has a length of its conjugate axis equal to and the distance between its foci is , then the eccentricity of the hyperbola is:
Q. If focii of x2a2−y2b2=1 coincide with the focii of x225+y29=1 and eccentricity of the hyperbola is 2, then
- a2+b2=14
- There is a director circle of the hyperbola
- Length of latus rectum of the hyperbola is 12
- Centre of the director circle is (0, 0)
Q.
Write the length of the latus-rectum of the hyperbola16x2−9y2=144
Q. Let a+b+c=10, a, b, c>0. Then the maximum value of a2b3c5 is
- (15)3×100
- (15)3×10
- (15)3×1000
- (60)3
Q. The solution of dydx=x2+y2+12xy satisfying y(1) = 0 is
- an ellipse with e =1√2
- a hyperbola with e = √2
- a circle whose radius is 1
- a curve symmetric about origin
Q. A variable plane forms a tetrahedron of constant volume 64K3 with the coordinate planes and the origin. The locus of the centroid of the tetrahedron is
- xyz=6k3
- x3+y3+z3=6K2
- x2+y2+z2=4K2
- x−2+y−2+z−2=4k−2
Q. A plane intersects the co ordinate axes at A, B, C. If O=(0, 0, 0) and (1, 1, 1) is the centroid of the tetrahedron OABC, then the sum of the reciprocals of the intercepts of the plane
- 12
- 43
- 1
- 34
Q. The hyperbola x2a2−y2b2=1 has its conjugate axis of length 5 and passes through the point (2, 1). The length of latus rectum is :
- 54√29
- 58√29
- √294
- √29
Q. Find the vertex, axis, focus, directrix and latus rectum of the parabola, also draw their rough sketches
4y2+12x−20y+67=0
4y2+12x−20y+67=0
Q.
The area (in square units) of the quadrilateral formed by the two pairs of lines
l2x2−m2y2−n(lx+my)=0
and l2x2−m2y2+n(lx−my)=0 is