AIEEE 2012 | Ellipse Question 111 | Mathematics

An ellipse is drawn by taking a diameter of thec circle $${\left( {x - 1} \right)^2} + {y^2} = 1$$ as its semi-minor axis and a diameter of the circle $${x^2} + {\left( {y - 2} \right)^2} = 4$$ is semi-major axis. If the centre of the ellipse is at the origin and its axes are the coordinate axes, then the equation of the ellipse is :

$$4{x^2} + {y^2} = 4$$

$${x^2} + 4{y^2} = 8$$

$$4{x^2} + {y^2} = 8$$

$${x^2} + 4{y^2} = 16$$

AIEEE 2011

MCQ (Single Correct Answer)

-1

Equation of the ellipse whose axes of coordinates and which passes through the point $$(-3,1)$$ and has eccentricity $$\sqrt {{2 \over 5}} $$ is :

$$5{x^2} + 3{y^2} - 48 = 0$$

$$3{x^2} + 5{y^2} - 15 = 0$$

$$5{x^2} + 3{y^2} - 32 = 0$$

$$3{x^2} + 5{y^2} - 32 = 0$$

AIEEE 2009

MCQ (Single Correct Answer)

-1

The ellipse $${x^2} + 4{y^2} = 4$$ is inscribed in a rectangle aligned with the coordinate axex, which in turn is inscribed in another ellipse that passes through the point $$(4,0)$$. Then the equation of the ellipse is :

$${x^2} + 12{y^2} = 16$$

$$4{x^2} + 48{y^2} = 48$$

$$4{x^2} + 64{y^2} = 48$$

$${x^2} + 16{y^2} = 16$$

AIEEE 2008

MCQ (Single Correct Answer)

-1

A focus of an ellipse is at the origin. The directrix is the line $$x=4$$ and the eccentricity is $${{1 \over 2}}$$. Then the length of the semi-major axis is :

$${{8 \over 3}}$$

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

$${{4 \over 3}}$$

$${{5 \over 3}}$$

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