1
JEE Main 2016 (Offline)
+4
-1
Out of Syllabus
Let $$P$$ be the point on the parabola, $${{y^2} = 8x}$$ which is at a minimum distance from the centre $$C$$ of the circle, $${x^2} + {\left( {y + 6} \right)^2} = 1$$. Then the equation of the circle, passing through $$C$$ and having its centre at $$P$$ is:
A
$${{x^2} + {y^2} - {x \over 4} + 2y - 24 = 0}$$
B
$${{x^2} + {y^2} - 4x + 9y + 18 = 0}$$
C
$${{x^2} + {y^2} - 4x + 8y + 12 = 0}$$
D
$${{x^2} + {y^2} - x + 4y - 12 = 0}$$
2
JEE Main 2015 (Offline)
+4
-1
Let $$O$$ be the vertex and $$Q$$ be any point on the parabola, $${{x^2} = 8y}$$. If the point $$P$$ divides the line segment $$OQ$$ internally in the ratio $$1:3$$, then locus of $$P$$ is :
A
$${y^2} = 2x$$
B
$${{x^2} = 2y}$$
C
$${{x^2} = y}$$
D
$${y^2} = x$$
3
JEE Main 2014 (Offline)
+4
-1
Out of Syllabus
The slope of the line touching both the parabolas $${y^2} = 4x$$ and $${x^2} = - 32y$$ is
A
$${{1 \over 8}}$$
B
$${{2 \over 3}}$$
C
$${{1 \over 2}}$$
D
$${{3 \over 2}}$$
4
JEE Main 2013 (Offline)
+4
-1
Out of Syllabus
Given : A circle, $$2{x^2} + 2{y^2} = 5$$ and a parabola, $${y^2} = 4\sqrt 5 x$$.
Statement-1 : An equation of a common tangent to these curves is $$y = x + \sqrt 5$$.

Statement-2 : If the line, $$y = mx + {{\sqrt 5 } \over m}\left( {m \ne 0} \right)$$ is their common tangent, then $$m$$ satiesfies $${m^4} - 3{m^2} + 2 = 0$$.

A
Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
B
Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
C
Statement-1 is true; Statement-2 is false.
D
Statement-1 is false Statement-2 is true.
EXAM MAP
Medical
NEET