Maths-
General
Easy
Question
The radius of the circle passing through the points of intersection of ellipse 
= 1 and x2 – y2 = 0 is -
The correct answer is:
Two curves are symmetrical about both axes and intersect in four points, so, the circle through their points of intersection will have centre at origin.
Solving 
= 0 and = 1, we get
= 
Therefore radius of circle
=
=
Related Questions to study
maths-
If 
are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -
are collinear.
e =
= 
= 
If are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -
maths-General
are collinear. e = = = maths-
If 
is the angle between the diameter through any point on a standard ellipse and the normal at the point, then the greatest value of tan is–
Any point P on ellipse is (a cos 
, b sin )
Equation of the diameter CP is y =
x
The normal to ellipse at P is
ax sec – by cosec = a2e2
Slopes of the lines CP and the normal GP are
tan and
tan
tan = 
=
=
sin cos = 
sin 2
The greatest value of tan = .1 = .
, b sin ) Equation of the diameter CP is y =
xThe normal to ellipse at P is
ax sec
– by cosec = a2e2Slopes of the lines CP and the normal GP are
tan andtan tan
= ==
sin cos = sin 2 The greatest value of tan
= .1 = .If is the angle between the diameter through any point on a standard ellipse and the normal at the point, then the greatest value of tan is–
maths-General
Any point P on ellipse is (a cos , b sin )
Equation of the diameter CP is y =x
The normal to ellipse at P is
ax sec – by cosec = a2e2
Slopes of the lines CP and the normal GP aretan andtan
tan = =
=sin cos = sin 2
The greatest value of tan = .1 = .
, b sin ) Equation of the diameter CP is y =
xThe normal to ellipse at P is
ax sec
– by cosec = a2e2Slopes of the lines CP and the normal GP are
tan andtan tan
= ==
sin cos = sin 2 The greatest value of tan
= .1 = .physics-
A radar operates at wavelength 50.0 cm. If the beat freqency between the transmitted singal and the singal reflected from aircraft (
) is 1 kHz, then velocity of the aircraft will be
when source is fixed and observer is moving towards it
when source is moving towards observer at rest
= 900 km/hr
when source is moving towards observer at rest
= 900 km/hr
A radar operates at wavelength 50.0 cm. If the beat freqency between the transmitted singal and the singal reflected from aircraft ( ) is 1 kHz, then velocity of the aircraft will be
physics-General
when source is fixed and observer is moving towards it
when source is moving towards observer at rest
= 900 km/hr
when source is moving towards observer at rest
= 900 km/hr
Maths-
The locus of P such that PA2 + PB2 = 10 where A = (2, 0) and B = (4, 0) is
The locus of P such that PA2 + PB2 = 10 where A = (2, 0) and B = (4, 0) is
Maths-General
maths-
Let L = 0 is a tangent to ellipse 
+ 
= 1 and S, 
be its foci. If length of perpendicular from S on L = 0 is 2 then length of perpendicular from on L = 0 is
Let L = 0 is a tangent to ellipse + = 1 and S, be its foci. If length of perpendicular from S on L = 0 is 2 then length of perpendicular from on L = 0 is
maths-General
maths-
The condition that the line 
x + my = n may be a normal to the ellipse 
+ 
= 1 is
The condition that the line x + my = n may be a normal to the ellipse + = 1 is
maths-General
Physics-
Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lie on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a (
). (g is acceleration due to gravity). Normal reaction (in N) acting on block A.
Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lie on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a ( ). (g is acceleration due to gravity). Normal reaction (in N) acting on block A.
Physics-General
physics-
Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lie on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a ( ). (g is acceleration due to gravity) Normal reaction (in N) acting on block B is
Two blocks A and B of equal masses m kg each are connected by a light thread, which passes over a massless pulley as shown in the figure. Both the blocks lie on wedge of mass m kg. Assume friction to be absent everywhere and both the blocks to be always in contact with the wedge. The wedge lying over smooth horizontal surface is pulled towards right with constant acceleration a ( ). (g is acceleration due to gravity) Normal reaction (in N) acting on block B is
physics-General
physics-
A light inextensible string connects a block of mass m and top of wedge of mass M. The string is parallel to inclined surface and the inclined surface makes an angle with horizontal as shown in the figure. All surfaces are smooth. Now a constant horizontal force of minimum magnitude F is applied to wedge towards right such that the normal reaction on block exerted by wedge just becomes zero. The magnitude of acceleration of wedge is
A light inextensible string connects a block of mass m and top of wedge of mass M. The string is parallel to inclined surface and the inclined surface makes an angle with horizontal as shown in the figure. All surfaces are smooth. Now a constant horizontal force of minimum magnitude F is applied to wedge towards right such that the normal reaction on block exerted by wedge just becomes zero. The magnitude of acceleration of wedge is
physics-General
maths-
The vector 
which is perpendicular to (2,-3,1) and (1,-2,3) and which satisfies the condition 
The vector which is perpendicular to (2,-3,1) and (1,-2,3) and which satisfies the condition
maths-General
Maths-
Let f : R → R be a differentiable function satisfying f (x) = f (x – y) f (y) " x, y Î R and f ¢ (0) = a, f ¢ (2) = b then f ¢ (-2) is
Let f : R → R be a differentiable function satisfying f (x) = f (x – y) f (y) " x, y Î R and f ¢ (0) = a, f ¢ (2) = b then f ¢ (-2) is
Maths-General
physics-
Four identical metal butterflies are hanging from a light string of length 
at equally placed points as shown in the figure . The ends of the string are attached to a horizontal fixed support. The middle section of the string is horizontal. The relation between the angle 
and 
is given by
Four identical metal butterflies are hanging from a light string of length at equally placed points as shown in the figure . The ends of the string are attached to a horizontal fixed support. The middle section of the string is horizontal. The relation between the angle and is given by
physics-General
physics-
Initially the spring is undeformed. Now the force 'F' is applied to 'B' as shown in the figure . When the displacement of 'B' w.r.t 'A' is 'x' towards right in some time then the relative acceleration of 'B' w.r.t. 'A' at that moment is:
F - Kx = mb and kx = ma
Hence m (b – a) = F – 2kx
Hence m (b – a) = F – 2kx
Initially the spring is undeformed. Now the force 'F' is applied to 'B' as shown in the figure . When the displacement of 'B' w.r.t 'A' is 'x' towards right in some time then the relative acceleration of 'B' w.r.t. 'A' at that moment is:
physics-General
F - Kx = mb and kx = ma
Hence m (b – a) = F – 2kx
Hence m (b – a) = F – 2kx
physics-
Mass m shown in the figure is in equilibrium. If it is displaced further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light.
Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx
\ acceleration of the block is = m/4kx
\ acceleration of the block is = m/4kx
Mass m shown in the figure is in equilibrium. If it is displaced further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light.
physics-General
Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx
\ acceleration of the block is = m/4kx
\ acceleration of the block is = m/4kx
physics-
In the figure shown, the pulleys and strings are massless. The acceleration of the block of mass 4m just after the system is released from rest is 
The FBD of blocks is as shown From Newton's second law 4mg – 2T cosq = 4 mA .... (1)
and T – mg = ma .... (2)
cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g
and T – mg = ma .... (2)
cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g
In the figure shown, the pulleys and strings are massless. The acceleration of the block of mass 4m just after the system is released from rest is
physics-General
The FBD of blocks is as shown From Newton's second law 4mg – 2T cosq = 4 mA .... (1)
and T – mg = ma .... (2)
cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g
and T – mg = ma .... (2)
cosq = 5/4 and from constraint we get a = A cosq (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 11/5g
