The equation of the auxiliary circle of x2/16-y2/25=1 is
x2+y2=16
x2+y2=9
x2+y2=5
x2+y2=15
If the normal at ‘θ’ on the hyperbola x2/a2-y2/b2=1 meets the tansverse axis at G, the AG, AG’=
a2(e4 sec2 θ-1)
a2(e4 sec2 θ+1)
b2(e4 sec2 θ-1)
none
The equations to the common tangents to the two hyperbolas x2/a2-y2/b2=1 and y2/a2-x2/b2=1Are
y= ±x±√b2-a2
y= ±x±√a2-b2
y= ±x±(a2-b2)
y= ±x±√a2+b2
The conic represented by 2x2-12xy+23y2-4x-28y-48=0 is
parabola
Ellipse
hyperbola
A line through the origin meets the circle x2+y2=a2 at P and the hyperbola x2-y2=a2 at Q. The locus the point of intersection of the tangent at P to the circle and with the tangent t Q to the hyperbola is
(x4+y4)=a6
(a4+4y4) x2=a6
(a4+4x4) y2=a6
Tangents are drawn from the Point (-2, -1) to the hyperbola 2x2-3y2=6. Their equations are
3x-y+5=0, x-y+1=0
3x+y+5=0, x+y+1=0
3x-y-5=0, x-y-1=0
3x+y-5=0, x+y-1=0
The equation of the hyperbola whose eccentricity 2 and foci are the foci of the ellipse x2/25 +y2/9 =1 is
x2/4--y2/12=1
x2/14+y2/12=1
x2/15+y2/8=1
x2/6-y2/8=1
A normal to the hyperbola x2/a2-y2/b2=1 cuts the axes at K and L. The perpendiculars at K and L axes meet in P. The locus of P is
a2x2+b2y2=(a2+b2)2
a2x2-b2y2=(a2+b2)2
a2x2+b2y2=(a2-b2)2
a2x2-b2y2=(a2-b2)2
For a binominal variate X, if n = 4 and P(X = 4) = 6 P(X = 2), then the value of p is:
3/7
4/7
6/7
5/7
The curve represented by x=a(cosh θ+sinh θ), y=b(cosh θ-sinh θ) is
a hyperbola
an ellipse
a parabola
a circle
The locus of poles of the lines with respect to the hyperbola x2/a2-y2/b2=1 which touch the ellipse x2/α2+y2/β2=1is
α2 x2/a4+ β2 y2/b4=1
α2 x2/a4-β2 y2/b4=1
α2 x2/a4+ α2 y2/b4=1
α2 x2/a2+ β2 y2/b2=1
The condition that the line x cos α + y sin α =p to be a tangent to the hyperbola x2/a2 -y2/b2 =1 is
a2 cos2 α+ b2 sin2 α =p2
a2 cos2 α- b2 sin2 α =p2
a2 sin2 α+ b2 cos2 α =p2
If the asymptotes of the hyperbola 14x2+38xy+20y2+x-7y-91=0 are 7x+5y-3=0, ax+by+c=0 then the descending order of a, b, c is
a, b, c
b, c, a
c, a, b
c, b, a
The distance between the foci of the hyperbola x2- 3y2- 4x - 6y - 11 = 0 is
4
6
8
10
If m1, m2 are slopes of the tangents to the hyperbola x2/25-y2/16=1 which pass through the point (6, 2) then
m1+ m2=24/11
m1+ m2=48/11
m1m2=28/11
m1 m2=11/20
The equation of the hyperbola which passes through the point (2,3) and has the asymptotes 4x+3y-7=0 and x-2y-1=0 is
4x2+5xy-6y2-11x+11y+50=0
4x2+5xy-6y2-11x+11y-43=0
4x2-5xy-6y2-11x+11y+57=0
x2-5xy-y2-11x+11y-43=0
The point of contact of 5x+6y+1=0 to the hyperbola 2x2-3y2 =2 is
(3,4)
(-5,4)
(-5,-4)
(5,-4)
The centre of the hyperbola 9x2-16y2+72x-32y-16=0 is
(1,1)
(1,-1)
(-1,1)
(-4,-1)
The foot of the normal 3x+4y=7 to the hyperbola 4x2-3y2=1 is
(1, 1)
(1, -1)
(-1, 1)
(-1, -1)
The locus of poles of tangents to the circle x2+y2=a2-b2 w. r. t the hyperbola x2/a2-y2/b2=1is
x2/a4+y2/b4=1/a2-b2
x2/a4-y2/b4=1/a2-b2
x2/a4+y2/b4=1/a2+b2
x2/a4-y2/b4=1/a2+b2
The equation of the hyperbola with its transverse axis is parallel to y-axis, and its centre is (2,-3), the length of transverse axis is 12 and eccentricity 7/6 is
(y+3)2/6-(x+2)2/3=1
(y-3)2/36-(x-2)2/13=1
(y+3)2/9-(x-2)2/4=1
(y+3)2/36-(x-2)2/13=1
The equation of the transverse and conjugate axes of a hyperbola are respectively. X+2y-3=0, 2x-y+4=0 and their respective lengths are √2 and 2/√3. The equation of the hyperbola is
2/5(x+2y-3)2-3/5(2x-y+4)2 =1
2/5(2x+y+3)2-3/5(x+2y-3)2 =1
2(2x-y+4)2-3(x+2y-3)2 =1
2(x+2y-3)2-3(2x-y+4)2 =1
The length of the latus rectum of the hyperbola 9x2-16y2+72x-32y-16=0 is
9/2
32/3
11/5
21/5
The sum and product of the slops of the tangents to the hyperbola 2x2-3y2=6 drawn from the point (-1,1) are
1,-3
1,-3/2
2, -3/2
3,-2/2
The equation of the normal at the positive end of the latus rectum of the hyperbola x2-3y2=144 is
√3x+2y=32
√3x-3y=48
3x+√3y=48
3x-√3y=48
The length of latus rectum of parabola y2+8x-2y+17 = 0 is:
2
16
Radius of the director circle of the hyperbola (x2/81) - (y2/36) = 1 is
2√5
√5
3√5
√5/2
If a hyperbola has one focus at the origin and its eccentricity is √2. One of the directries is x+y+1=0. Then the equations to its asymptotes are
x-1=0, y-1=0
x+1=0, y+1=0
x+3=0, y+3=0
x+2=0, y+2=0
The equation of the normal to the hyperbola x2-4y2= 5 at (3,-1) is
4x+3y-15=0
4x-3y+15=0
4x-3y+5=0
4x+4y+15=0
The foci of the hyperbola 2x2-y2-4x+4y-10=0 are
(±√13, 0)
(1±2√3, 2)
(2±3√3, 3)
(3±3√3, 2)