JEE Main 2019 (Online) 11th January Morning Slot | Conic Sections Question 170 | Mathematics

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JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)

-1

Equation of a common tangent to the parabola y² = 4x and the hyperbola xy = 2 is

x + y + 1 = 0

4x + 2y + 1 = 0

x – 2y + 4 = 0

x + 2y + 4 = 0

JEE Main 2019 (Online) 11th January Morning Slot

MCQ (Single Correct Answer)

-1

If tangents are drawn to the ellipse x2 + 2y² = 2 at all points on the ellipse other than its four vertices then the mid points of the tangents intercepted between the coordinate axes lie on the curve :

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

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

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

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

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

The length of the chord of the parabola x² $$=$$ 4y having equation x – $$\sqrt 2 y + 4\sqrt 2 = 0$$ is -

$$8\sqrt 2 $$

$$6\sqrt 3 $$

$$3\sqrt 2 $$

$$2\sqrt {11} $$

JEE Main 2019 (Online) 10th January Evening Slot

MCQ (Single Correct Answer)

-1

Let S = $$\left\{ {\left( {x,y} \right) \in {R^2}:{{{y^2}} \over {1 + r}} - {{{x^2}} \over {1 - r}}} \right\};r \ne \pm 1.$$ Then S represents

an ellipse whose eccentricity is $${1 \over {\sqrt {r + 1} }},$$ where r > 1

an ellipse whose eccentricity is $${2 \over {\sqrt {r + 1} }},$$ where 0 < r < 1

an ellipse whose eccentricity is $${2 \over {\sqrt {r - 1} }},$$ where 0 < r < 1

an ellipse whose eccentricity is $$\sqrt {{2 \over {r + 1}}}$$, where r > 1

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