IIT-JEE 2006 | Parabola Question 41 | Mathematics

IIT-JEE 2006

MCQ (Single Correct Answer)

-0.75

The axis of a parabola is along the line $$y = x$$ and the distances of its vertex and focus from origin are $$\sqrt 2 $$ and $$2\sqrt 2 $$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is

$${\left( {x + y} \right)^2} = \left( {x - y - 2} \right)$$

$${\left( {x - y} \right)^2} = \left( {x + y - 2} \right)$$

$${\left( {x - y} \right)^2} = 4\left( {x + y - 2} \right)$$

$${\left( {x - y} \right)^2} = 8\left( {x + y - 2} \right)$$

IIT-JEE 2006

MCQ (Single Correct Answer)

-0

$$ \text { Normals are drawn at points } \mathrm{P}, \mathrm{Q} \text { and } \mathrm{R} \text { lying on the parabola } y^2=4 x \text { which intersect at }(3,0) \text {. Then } $$

(i)	Area of $\triangle \mathrm{PQR}$	(A)	2
(ii)	Radius of circumcircle of $\triangle \mathrm{PQR}$	(B)	5/2
(iii)	Centroid of $\triangle \mathrm{PQR}$	(C)	(5/2,0)
(iv)	Circumcentre of $\triangle \mathrm{PQR}$	(D)	(2/3,0)

$$ \begin{aligned} & \text { (i)-(A); (ii)-(B); (iii)-(D); } \text { (iv)-(C) } \end{aligned} $$

$$ \begin{aligned} & \text { (i)-(B); (ii)-(A); (iii)-(D); } \text { (iv)-(C) } \end{aligned} $$

$$ \begin{aligned} & \text { (i)-(A); (ii)-(B); (iii)-(C); } \text { (iv)-(D) } \end{aligned} $$

$$ \begin{aligned} & \text { (i)-(A); (ii)-(D); (iii)-(B); } \text { (iv)-(C) } \end{aligned} $$

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)

-0.5

Tangent to the curve $$y = {x^2} + 6$$ at a point $$(1, 7)$$ touches the circle $${x^2} + {y^2} + 16x + 12y + c = 0$$ at a point $$Q$$. Then the coordinates of $$Q$$ are

$$(-6, -11)$$

$$(-9, -13)$$

$$(-10, -15)$$

$$(-6, -7)$$

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.5

The angle between the tangents drawn from the point $$(1, 4)$$ to the parabola $${y^2} = 4x$$ is

$$\pi /6$$

$$\pi /4$$

$$\pi /3$$

$$\pi /2$$

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