Eccentricity of Hyperbola
Trending Questions
The locus of the vertices of the family of parabolas is
Parametric equation of rectangular hyperbola xy = c2 is x = ct & y = ct where t ϵ R.
False
True
The shortest distance from the point to the surface of the sphere is
Find the equation of the hyperbola whose vertices are (±7, 0) and the eccentricity is 43.
The …… of a conic is the chord passing through the focus and perpendicular to the axis.
Line of axis
Directrix
Y-axis
Latus-rectum
The coordinates of the focus of the parabola described parametrically by are:
Find the coordinates of the foci, the vertices, the eccentricity and the length of the latus rectum of the hyperbola,
5y2−9x2=36
If e1 is the eccentricity of the ellipse x216+y225=1 and e2 is the eccentricity of the hyperbola passing through the foci of the ellipse and e1e2=1, then equation of the hyperbola is
x29−y216=1
x216−y29=−1
x29−y225=1
None of these
Let P(6, 3) be a point on hyperbola x2a2−y2b2=1. If the normal at the point P intersect the x - axis at (9, 0), then the eccentricity of the hyperbola is
(IIT JEE 2011)
√52
√32
√3
√2
e and e1 are the eccentricities of the hyperbolas 16x2−9y2=144 and 9x2−16y2= - 144 then e - e1 =
2
1
0
32
- its length of transverse axis is 8 units
- its length of conjugate axis is 2√3 units
- end points of one latus rectum will be (14, 2+√19) and (74, 2+√19)
- Distance between the two directrices is 16√19 units
(i) 16x2 − 9y2 + 32x + 36y − 164 = 0
(ii) x2 − y2 + 4x = 0
(iii) x2 − 3y2 − 2x = 8.
e1 and e2 are the eccentricities of two conics S and S1. If e12+e22 = 3 then both S and S1 can be
ellipse
parabolas
hyperbolas
circles
- 2√3
- √3√2
- √2
- √133
- (±2, 0)
- (±2√3, 0)
- (±2√3, 0)
- (±23, 0)
- 1
- 4
- 3
- 2