1
JEE Main 2015 (Offline)
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 5} = 1$$, is :
A
$${{27 \over 2}}$$
B
$$27$$
C
$${{27 \over 4}}$$
D
$$18$$
2
JEE Main 2014 (Offline)
MCQ (Single Correct Answer)
+4
-1
The locus of the foot of perpendicular drawn from the centre of the ellipse $${x^2} + 3{y^2} = 6$$ on any tangent to it is :
A
$$\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} + 2{y^2}$$
B
$$\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} - 2{y^2}$$
C
$$\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} + 2{y^2}$$
D
$$\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} - 2{y^2}$$
3
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
$${x^2} + {y^2} - 6y + 5 = 0$$
4
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.
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