JEE Main 2015 (Offline) | Conic Sections Question 228 | Mathematics

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)

-1

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

$${{27 \over 2}}$$

$$27$$

$${{27 \over 4}}$$

$$18$$

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

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

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

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

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

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