The abscissa of a point on the curve `x y=(a+x)^2,` the normal which cuts off numerically equal intercepts from the coordinate axes, is (a) `-1/(sqrt(2))` (b) `sqrt(2)a` (c) `a/(sqrt(2))` (d) `-sqrt(2)a`

The abscissa of a point on the curve x y=(a+x)^2, the normal which cuts off numerically equal intercepts from the coordinate axes, is -1/(sqrt(2)) (b) sqrt(2)a (c) a/(sqrt(2)) (d) -sqrt(2)a

The abscissa of the point on the curve ay^2 = x^3 , the normal at which cuts off equal intercepts from the coordinate axes is

The abscissa of the point on the curve sqrt(x y)=a+x the tangent at which cuts off equal intercepts from the coordinate axes is -a/(sqrt(2)) (b) a//sqrt(2) (c) -asqrt(2) (d) asqrt(2)

The point on the curve sqrt(x) + sqrt(y) = sqrt(a) , the normal at which is parallel to the x-axis, is

int_0^(sqrt(2)) [x^2] dx is equal to (A) 2-sqrt(2) (B) 2+sqrt(2) (C) sqrt(2)-1 (D) sqrt(2)-2

The length of the chord of the parabola y^2=x which is bisected at the point (2, 1) is (a) 2sqrt(3) (b) 4sqrt(3) (c) 3sqrt(2) (d) 2sqrt(5)

1/(sqrt(9)-\ sqrt(8)) is equal to: (a) 3+2sqrt(2) (b) 1/(3+2sqrt(2)) (c) 3-2sqrt(2) (d) 3/2-\ sqrt(2)

The area bounded by the coordinate axes and the curve sqrt(x) + sqrt(y) = 1 , is

A(1/(sqrt(2)),1/(sqrt(2))) is a point on the circle x^2+y^2=1 and B is another point on the circle such that are length A B=pi/2 units. Then, the coordinates of B can be (a) (1/(sqrt(2)),-1/sqrt(2)) (b) (-1/(sqrt(2)),1/sqrt(2)) (c) (-1/(sqrt(2)),-1/(sqrt(2))) (d) none of these

Radius of the circle that passes through the origin and touches the parabola y^2=4a x at the point (a ,2a) is (a) 5/(sqrt(2))a (b) 2sqrt(2)a (c) sqrt(5/2)a (d) 3/(sqrt(2))a