Let T and C respectively be the transverse and conjugate axes of the hyperbola $$16x^2-y^2+64x+4y+44=0$$. Then the area of the region above the parabola $$x^2=y+4$$, below the transverse axis T and on the right of the conjugate axis C is :

Let R be a rectangle given by the lines $$x=0,x=2,y=0$$ and $$y=5$$. Let A$$(\alpha ,0)$$ and B$$(0,\beta ),\alpha \in [0,2]$$ and $$\beta \in [0,5]$$, be such that the line segment AB divides the area of the rectangle R in the ratio 4 : 1. Then, the mid-point of AB lies on a :

For a triangle $$ABC$$, the value of $$\cos 2A+\cos 2B+\cos 2C$$ is least. If its inradius is 3 and incentre is M, then which of the following is NOT correct?

A straight line cuts off the intercepts $$\mathrm{OA}=\mathrm{a}$$ and $$\mathrm{OB}=\mathrm{b}$$ on the positive directions of $$x$$-axis and $$y$$ axis respectively. If the perpendicular from origin $$O$$ to this line makes an angle of $$\frac{\pi }{6}$$ with positive direction of $$y$$-axis and the area of $$△\mathrm{OAB}$$ is $$\frac{98}{3}\sqrt{3}$$, then $${\mathrm{a}}^2-{\mathrm{b}}^2$$ is equal to :

In a triangle ABC, if $$\cos \mathrm{A}+2\cos \mathrm{B}+\cos C=2$$ and the lengths of the sides opposite to the angles A and C are 3 and 7 respectively, then $$\cos \mathrm{A}-\cos \mathrm{C}$$ is equal to

Let the plane P pass through the intersection of the planes $$2x+3y-z=2$$ and $$x+2y+3z=6$$, and be perpendicular to the plane $$2x+y-z+1=0$$. If d is the distance of P from the point ($$-$$7, 1, 1), then $${\mathrm{d}}^2$$ is equal to :

Let the plane $\mathrm{P}:8x+{\alpha }_{1}y+{\alpha }_{2}z+12=0$ be parallel to the line $\mathrm{L}:\frac{x+2}{2}=\frac{y-3}{3}=\frac{z+4}{5}$. If the intercept of $\mathrm{P}$ on the $y$-axis is 1 , then the distance between $\mathrm{P}$ and $\mathrm{L}$ is :

The foot of perpendicular from the origin $\mathrm{O}$ to a plane $\mathrm{P}$ which meets the co-ordinate axes at the points $\mathrm{A},\mathrm{B},\mathrm{C}$ is $(2,\mathrm{a},4),\mathrm{a}\in \mathrm{N}$. If the volume of the tetrahedron $\mathrm{OABC}$ is 144 unit${}^3$, then which of the following points is NOT on P ?

Let $P$ be the plane, passing through the point $(1,-1,-5)$ and perpendicular to the line joining the points $(4,1,-3)$ and $(2,4,3)$. Then the distance of $P$ from the point $(3,-2,2)$ is :

If a point $\mathrm{P}(\alpha ,\beta ,\gamma )$ satisfying \left( {\matrix{ \alpha & \beta & \gamma \cr } } \right)\left( {\matrix{ 2 & {10} & 8 \cr 9 & 3 & 8 \cr 8 & 4 & 8 \cr } } \right) = \left( {\matrix{ 0 & 0 & 0 \cr } } \right) lies on the plane $2 x+4 y+3 z=5$, then $6\alpha +9\beta +7\gamma$ is equal to :

The shortest distance between the lines {{x - 5} \over 1} = {{y - 2} \over 2} = {{z - 4} \over { - 3}} and {{x + 3} \over 1} = {{y + 5} \over 4} = {{z - 1} \over { - 5}} is :

Let the image of the point $$P(2,-1,3)$$ in the plane $$x+2y-z=0$$ be $$Q$$. Then the distance of the plane $$3x+2y+z+29=0$$ from the point $$Q$$ is :

Let the shortest distance between the lines $$L:\frac{x-5}{-2}=\frac{y-\lambda }{0}=\frac{z+\lambda }{1},\lambda \ge 0$$ and $$L_1:x+1=y-1=4-z$$ be $$2\sqrt{6}$$. If $$(\alpha ,\beta ,\gamma )$$ lies on $$L$$, then which of the following is NOT possible?

A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\circ }$, to the $y$-axis at 45 and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then :

If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is :

The line $$l_1$$ passes through the point (2, 6, 2) and is perpendicular to the plane $$2x+y-2z=10$$. Then the shortest distance between the line $$l_1$$ and the line $$\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$$ is :

The plane $$2x-y+z=4$$ intersects the line segment joining the points A ($$a,-2,4)$$ and B ($$2,b,-3)$$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $$\sqrt{5}$$. If $$ab

If the lines {{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1} and {{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1} intersect at the point P, then the distance of the point P from the plane $$z=a$$ is :

The shortest distance between the lines {{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5} and {{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}} is :

The foot of perpendicular of the point (2, 0, 5) on the line {{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}} is ($$\alpha ,\beta ,\gamma$$). Then, which of the following is NOT correct?