Parametric Form of Normal : Hyperbola
Trending Questions
The equation of normal to a hyperbola x2a2−y2b2=1 at a point with parameter θ is
axcosθ−bycotθ=a2+b2
Differential coefficient of with respect to at is
Difference of slopes of the lines represented by equation is
none of these
- a : b
- a2 : b2
- b2 : a2
- b : a
- eccentricity equal to √3
- length of latus rectum equal to 20
- equation of auxiliary circle equal to x2+y2=25
- distance between foci equal to 5√5
- a2l2−b2m2=(a2+b2)2n2
- l2a2−m2b2=(a2+b2)2n2
- a2l2+b2m2=(a2−b2)2n2
- l2b2+m2b2=(a2−b2)2n2
- ±4acotα cosec α
- 4acotα cosec α
- −4acotα cosec α
- 4a cosec2 α
If and be the distances of origin from the lines and, then
- a2+b2a
- −(a2+b2a)
- a2+b2b
- −(a2+b2b)
Compute the area of the curvilinear triangle bounded by the -axis and the curves, and .
- ±32
- ±52
- None of these
- ±45
If and are the lengths of the subtangent and the subnormal at the point on the curve then
- PF⋅PG=CB2
- PF⋅PG=2CB2
- Locus of mid-point of GH is another hyperbola with eccentricity =√10
- PF⋅PH=CA2
- True
- False
A normal is drawn on the hyperbola x216−y29=1 at a point given by parameter π3.
What is the x-intercept of the normal.
What is the eccentricity of a hyperbola if a chord joining 2 points given by parameters α and β also passes through the focus. [Given :α + β = p, α − β = q]
cos
cos
- a2l2−b2m2=(a2+b2)2n2
- l2a2−m2b2=(a2+b2)2n2
- a2l2+b2m2=(a2−b2)2n2
- l2b2+m2b2=(a2−b2)2n2
- a2+b2a
- −(a2+b2a)
- a2+b2b
- −(a2+b2b)
- True
- False
- y=f(x)=x24−4
- minimum value of f(x) is −4
- Number of integral values of x lying between the roots of f(x) = 0 are 7
- Area of the △ABC is 16 sq.units
- y−2x−3=0
- y−2x+1=0
- y=x+1
- y=x+3
- 25
- 5
- 16
- None of these
- _v=−_b+_b×_c
- _v=34(_b+2_b×_c)
- none of these
- _v=14(_b+_b×_c)
Find the equation and the length of the common tangents to hyperbola x2a2−y2b2=1 and y2a2−x2b2=1
- PF⋅PG=CB2
- PF⋅PG=2CB2
- Locus of mid-point of GH is another hyperbola with eccentricity =√10
- PF⋅PH=CA2
- PG=PC
- Pg=PC
- PG=Pg
- Gg=2PC
- 30∘
- 45∘
- 60∘
- 90∘