JEE Main 2016 (Online) 9th April Morning Slot | Ellipse Question 72 | Mathematics

If the tangent at a point on the ellipse $${{{x^2}} \over {27}} + {{{y^2}} \over 3} = 1$$ meets the coordinate axes at A and B, and O is the origin, then the minimum area (in sq. units) of the triangle OAB is :

$${9 \over 2}$$

$$3\sqrt 3 $$

$$9\sqrt 3 $$

JEE Main 2015 (Offline)

MCQ (Single Correct Answer)

-1

Out of Syllabus

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 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 :

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

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

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

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

PREVIOUS NEXT

Questions Asked from Ellipse (MCQ (Single Correct Answer))

Number in Brackets after Paper Indicates No. of Questions