Welcome to Ctanujit Institute of Statistics & Mathematics...An Initiative by 'ISI'ian



Urgent Notice:

Due to unavoidable reason, Ctanujit Classes will be closed from 1st November to 30th November 2018.
Postal Package will be available from 1st December 2018.

IIT JEE SOLVED PAPERS ON CO-ORDINATE GEOMETRY(2D)

Here is some solved questions from IIT-JEE papers




Q. [IIT-JEE 2001]The equation of the directrix of the parabola          y2 + 4y +4x + 2 = 0 is
(a) x = -1      (b) x = 1      (c) x = -3/2      (d) x = 3/2
Solution :(d) Equation of the given parabola is (y+2)2 = - 4(x – ½)
Let   x – ½ = X,   y + 2 = Y . Then the above equation reduces to        Y2 = -4X . Equation of directrix is x – ½ = 1.
Q. [IIT-JEE 1998] The number of values of c such that the straight line y = 4x + c touched the curve   x2/4 + y2 = 1 is
(a) 0     (b) 1     (c) 2     (d) infinite
Solution :(c) The line y = mx + c touches the ellipse                      x2/a2 + y2/b2 =1 if c = ± √(a2m2+b2).
Thus, number of values of c is 2.
Q. [IIT-JEE 2002] The equation of the common tangent to the curves  y2 = 8x and  xy = -1 is
(a) 3y=9x+2    (b) y=2x+1     (c) 2y=x+8     (d) y=x+2
Solution :(d) Let equation of tangent to parabola  y2 = 8x be               y = mx + 2/m --------------(1)
Let (1) meet the hyperbola  xy = -1 --------(2) at the point given by                             x( mx + 2/m )=-1     m2x2 + 2x + m = 0 -------------(3)
Since the equation (1) is tangent to (2), equation (3) must have equal roots, for which  D = 0  4 – 4m2 = 0  m = 1.
From (1), equation of the common tangent is  y = x + 2 .