IIT-JEE 2005 Screening | Ellipse Question 27 | Mathematics

IIT-JEE 2005 Screening

MCQ (Single Correct Answer)

-0.5

The minimum area of triangle formed by the tangent to the $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ and coordinate axes is

$$ab$$ sq. units

$${{{{a^2} + {b^2}} \over 2}}$$ sq. units

$${{{{\left( {a + b} \right)}^2}} \over 2}$$ sq. units

$${{{a^2} + ab + {b^2}} \over 3}$$ sq. units

IIT-JEE 2005 Mains

MCQ (Single Correct Answer)

-1

Find the equation of the common tangent in the first quadrant to the circle $$x^{2}+y^{2}=16$$ and the ellipse $$\frac{x^{2}}{25}+\frac{y^{2}}{4}=1$$. Also find the length of the intercept of the tangent between the coordinate axes.

$$\frac{14}{\sqrt5}$$

$$\frac{5}{\sqrt3}$$

$$\frac{14}{\sqrt3}$$

$$\frac{15}{\sqrt3}$$

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.5

If tangents are drawn to the ellipse $${x^2} + 2{y^2} = 2,$$ then the locus of the mid-point of the intercept made by the tangents between the coordinate axes is

$${1 \over {2{x^2}}} + {1 \over {4{y^2}}} = 1$$

$${1 \over {4{x^2}}} + {1 \over {2{y^2}}} = 1$$

$${{{x^2}} \over 2} + {{{y^2}} \over 4} = 1$$

$${{{x^2}} \over 4} + {{{y^2}} \over 2} = 1$$

IIT-JEE 2003 Screening

MCQ (Single Correct Answer)

-0.5

The area of the quadrilateral formed by the tangents at the end points of latus rectum to the ellipse $${{{x^2}} \over 9} + {{{y^2}} \over 5} = 1,$$ is

$$27/4$$ sq. units

$$9$$ sq. units

$$27/2$$ sq. units

$$27$$ sq. units

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