JEE Main 2014 (Offline) | Conic Sections Question 209 | Mathematics

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)

-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

$$\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} + 2{y^2}$$

$$\left( {{x^2} + {y^2}} \right) ^2 = 6{x^2} - 2{y^2}$$

$$\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} + 2{y^2}$$

$$\left( {{x^2} - {y^2}} \right) ^2 = 6{x^2} - 2{y^2}$$

JEE Main 2014 (Offline)

MCQ (Single Correct Answer)

-1

The slope of the line touching both the parabolas $${y^2} = 4x$$ and $${x^2} = - 32y$$ is

$${{1 \over 8}}$$

$${{2 \over 3}}$$

$${{1 \over 2}}$$

$${{3 \over 2}}$$

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)

-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

$${x^2} + {y^2} - 6y - 7 = 0$$

$${x^2} + {y^2} - 6y + 7 = 0$$

$${x^2} + {y^2} - 6y - 5 = 0$$

$${x^2} + {y^2} - 6y + 5 = 0$$

JEE Main 2013 (Offline)

MCQ (Single Correct Answer)

-1

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 $${m^4} - 3{m^2} + 2 = 0$$.

Statement-1 is true; Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Statement-1 is true; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Statement-1 is true; Statement-2 is false.

Statement-1 is false Statement-2 is true.

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