IIT-JEE 2008 Paper 1 Offline | JEE Advanced Year Wise Previous Years Questions

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F, respectively. The line PQ is given by the equation $$\sqrt 3 x\, + \,y\, - \,6 = 0$$ and the point D is $$\left( {{{3\,\sqrt 3 } \over 2},\,{3 \over 2}} \right)$$. Further, it is given that the origin and the centre of C are on the same side of the line PQ.

The equation of circle C is

$${\left( {x\, - 2\sqrt 3 \,} \right)^2} + {(y - 1)^2} = 1$$

$${\left( {x\, - 2\sqrt 3 \,} \right)^2} + {(y + {1 \over 2})^2} = 1$$

$${\left( {x\, - \sqrt 3 \,} \right)^2} + {(y + 1)^2} = 1$$

$${\left( {x\, - \sqrt 3 \,} \right)^2} + {(y - 1)^2} = 1$$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Points E and F are given by

$$\left( {{{\,\sqrt 3 } \over 2},\,{3 \over 2}} \right),\,\left( {\sqrt 3 ,\,0} \right)$$

$$\left( {{{\,\sqrt 3 } \over 2},\,{1 \over 2}} \right),\,\left( {\sqrt 3 ,\,0} \right)$$

$$\left( {{{\,\sqrt 3 } \over 2},\,{3 \over 2}} \right),\,\left( {{{\,\sqrt 3 } \over 2},\,{1 \over 2}} \right)$$

$$\left( {{{\,3} \over 2},\,{{\sqrt 3 } \over 2}} \right),\,\left( {{{\,\sqrt 3 } \over 2},\,{1 \over 2}} \right)$$

IIT-JEE 2008 Paper 1 Offline

MCQ (Single Correct Answer)

-1

Equations of the sides QR, RP are

$$y = {2 \over {\sqrt 3 }}\,x + \,1,\,\,y = \, - {2 \over {\sqrt 3 }}\,x - 1$$

$$y = {1 \over {\sqrt 3 }}\,x,\,\,y = \,0$$

$$y = {{\sqrt 3 } \over 2}\,x + \,1,\,\,y = \, - {{\sqrt 3 } \over 2}\,x - 1$$

$$y = \sqrt 3 \,x,\,\,y = \,0$$

IIT-JEE 2008 Paper 1 Offline

MCQ (More than One Correct Answer)

-2

Let $$P\left( {{x_1},{y_1}} \right)$$ and $$Q\left( {{x_2},{y_2}} \right),{y_1} < 0,{y_2} < 0,$$ be the end points of the latus rectum of the ellipse $${x^2} + 4{y^2} = 4.$$ The equations of parabolas with latus rectum $$PQ$$ are :

$${x^2} + 2\sqrt 3y = 3 + \sqrt 3 $$

$${x^2} - 2\sqrt 3y = 3 + \sqrt 3 $$

$${x^2} + 2\sqrt 3y = 3 - \sqrt 3 $$

$${x^2} - 2\sqrt 3 y = 3 - \sqrt 3 $$

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