The graph of the curve `x^2+y^2-2x y-8x-8y+32=0` falls wholly in the (a)first quadrant (b) second quadrant (c)third quadrant (d) none of these

From the point (2, 2) tangent are drawn to the hyperbola (x^2)/(16)-(y^2)/9=1. Then the point of contact lies in the (a)first quadrant (b) second quadrant (c)third quadrant (d) fourth quadrant

The focus of the parabola x^2 -2x +8y + 17=0 is :

Find the area of the region bounded by x^(2)= 4y, y = 2, y = 4 and the y-axis in the first quadrant.

Find the area bounded by y^2lt=4x ,x^2+y^2geq2x ,a n dxlt=y+2 in the first quadrant.

cos x =- 1/2, x lies in third quadrant.

If two vertices of a triangle are (0,2) and (4,3) and its orthocentre is (0,0) then the third vertex of the triangle lies in (a) I^(st) quadrant (b) 2^(nd)quadrant (c) 3^(rd)quadrant (d) 4^(th)quadrant

The normal to the curve, x^2+""2x y-3y^2=""0,""a t""(1,""1) : (1) does not meet the curve again (2) meets the curve again in the second quadrant (3) meets the curve again in the third quadrant (4) meets the curve again in the fourth quadrant

The area (in square units) bounded by the curves y=sqrt(x),2y-x+3=0, x-axis, and lying in the first quadrant is

Find the area of the region bounded by y^(2) = 9x, x = 2, x = 4 and the x-axis in the first quadrant.

Find the greatest value of x^2 y^3 , where x and y lie in the first quadrant on the line 3x+4y=5 .