1
JEE Main 2013 (Offline)
MCQ (Single Correct Answer)
+4
-1
The equation of the circle passing through the foci of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$, and having centre at $$(0,3)$$ is :
A
$${x^2} + {y^2} - 6y - 7 = 0$$
B
$${x^2} + {y^2} - 6y + 7 = 0$$
C
$${x^2} + {y^2} - 6y - 5 = 0$$
D
$${x^2} + {y^2} - 6y + 5 = 0$$
2
AIEEE 2012
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
STATEMENT-1 : An equation of a common tangent to the parabola $${y^2} = 16\sqrt 3 x$$ and the ellipse $$2{x^2} + {y^2} = 4$$ is $$y = 2x + 2\sqrt 3 $$

STATEMENT-2 :If line $$y = mx + {{4\sqrt 3 } \over m},\left( {m \ne 0} \right)$$ is a common tangent to the parabola $${y^2} = 16\sqrt {3x} $$and the ellipse $$2{x^2} + {y^2} = 4$$, then $$m$$ satisfies $${m^4} + 2{m^2} = 24$$

A
Statement-1 is false, Statement-2 is true.
B
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
C
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
D
Statement-1 is true, Statement-2 is false.
3
AIEEE 2012
MCQ (Single Correct Answer)
+4
-1
An ellipse is drawn by taking a diameter of thec circle $${\left( {x - 1} \right)^2} + {y^2} = 1$$ as its semi-minor axis and a diameter of the circle $${x^2} + {\left( {y - 2} \right)^2} = 4$$ is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :
A
$$4{x^2} + {y^2} = 4$$
B
$${x^2} + 4{y^2} = 8$$
C
$$4{x^2} + {y^2} = 8$$
D
$${x^2} + 4{y^2} = 16$$
4
AIEEE 2011
MCQ (Single Correct Answer)
+4
-1
Equation of the ellipse whose axes of coordinates and which passes through the point $$(-3,1)$$ and has eccentricity $$\sqrt {{2 \over 5}} $$ is :
A
$$5{x^2} + 3{y^2} - 48 = 0$$
B
$$3{x^2} + 5{y^2} - 15 = 0$$
C
$$5{x^2} + 3{y^2} - 32 = 0$$
D
$$3{x^2} + 5{y^2} - 32 = 0$$
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