IIT-JEE 2005 Screening | Conic Sections Question 74 | 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 Screening

MCQ (Single Correct Answer)

-0.5

Tangent to the curve $$y = {x^2} + 6$$ at a point $$(1, 7)$$ touches the circle $${x^2} + {y^2} + 16x + 12y + c = 0$$ at a point $$Q$$. Then the coordinates of $$Q$$ are

$$(-6, -11)$$

$$(-9, -13)$$

$$(-10, -15)$$

$$(-6, -7)$$

IIT-JEE 2004 Screening

MCQ (Single Correct Answer)

-0.5

The angle between the tangents drawn from the point $$(1, 4)$$ to the parabola $${y^2} = 4x$$ is

$$\pi /6$$

$$\pi /4$$

$$\pi /3$$

$$\pi /2$$

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

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

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

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

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

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