A line parallel to the straight line 2x – y = 0 is tangent to the hyperbola {{{x^2}} \over 4} - {{{y^2}} \over 2} = 1 at the point $$\left(x_1,y_1\right)$$. Then $$x_1^2+5y_1^2$$ is equal to :

If e1 and e2 are the eccentricities of the ellipse, {{{x^2}} \over {18}} + {{{y^2}} \over 4} = 1 and the hyperbola, {{{x^2}} \over 9} - {{{y^2}} \over 4} = 1 respectively and (e1, e2) is a point on the ellipse, 15x2 + 3y2 = k, then k is equal to :

If a hyperbola passes through the point P(10, 16) and it has vertices at (± 6, 0), then the equation of the normal to it at P is :

A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If $$\angle BAC=90^o$$ and area$$\left(\Delta ABC\right)=5\sqrt{5}$$ s units, then the abscissa of the vertex C is :

The plane which bisects the line joining, the points (4, –2, 3) and (2, 4, –1) at right angles also passes through the point :

The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line {x \over 2} = {y \over 3} = {z \over { - 6}} is :

If (a, b, c) is the image of the point (1, 2, -3) in the line {{x + 1} \over 2} = {{y - 3} \over { - 2}} = {z \over { - 1}}, then a + b + c is :

If for some $$\alpha$$ $$\in$$ R, the lines L1 : {{x + 1} \over 2} = {{y - 2} \over { - 1}} = {{z - 1} \over 1} and L2 : {{x + 2} \over \alpha } = {{y + 1} \over {5 - \alpha }} = {{z + 1} \over 1} are coplanar, then the line L2 passes through the point :

The shortest distance between the lines {{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1} and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :

A plane P meets the coordinate axes at A, B and C respectively. The centroid of $$\Delta$$ABC is given to be (1, 1, 2). Then the equation of the line through this centroid and perpendicular to the plane P is :

A plane passing through the point (3, 1, 1) contains two lines whose direction ratios are 1, –2, 2 and 2, 3, –1 respectively. If this plane also passes through the point ($$\alpha$$, –3, 5), then $$\alpha$$ is equal to:

The foot of the perpendicular drawn from the point (4, 2, 3) to the line joining the points (1, –2, 3) and (1, 1, 0) lies on the plane :

Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6). Then the image of R in the plane P is :

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

The plane passing through the points (1, 2, 1), (2, 1, 2) and parallel to the line, 2x = 3y, z = 1 also passes through the point :

The mirror image of the point (1, 2, 3) in a plane is \left( { - {7 \over 3}, - {4 \over 3}, - {1 \over 3}} \right). Which of the following points lies on this plane ?

Let X = {x $$\in$$ N : 1 $$\le$$ x $$\le$$ 17} and Y = {ax + b: x $$\in$$ X and a, b $$\in$$ R, a > 0}. If mean and variance of elements of Y are 17 and 216 respectively then a + b is equal to :

If $$\sum _{i=1}^n\left(x_i-a\right)=n$$ and $$\sum _{i=1}^n{\left(x_i-a\right)}^2=na$$ (n, a > 1) then the standard deviation of n observations x1 , x2 , ..., xn is :

If the mean and the standard deviation of the data 3, 5, 7, a, b are 5 and 2 respectively, then a and b are the roots of the equation :

The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :