WB JEE 2023 | Ellipse and Hyperbola Question 15 | Mathematics

WB JEE 2023

MCQ (More than One Correct Answer)

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Let f be a strictly decreasing function defined on R such that $$f(x) > 0,\forall x \in R$$. Let $${{{x^2}} \over {f({a^2} + 5a + 3)}} + {{{y^2}} \over {f(a + 15)}} = 1$$ be an ellipse with major axis along the y-axis. The value of 'a' can lie in the interval (s)

$$( - \infty , - 6)$$

$$( - 6,2)$$

$$(2,\infty )$$

$$( - \infty ,\infty )$$

WB JEE 2022

MCQ (More than One Correct Answer)

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Chords of an ellipse are drawn through the positive end of the minor axis. Their midpoint lies on

a circle

a parabola

an ellipse

a hyperbola

WB JEE 2022

MCQ (More than One Correct Answer)

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The line y = x + 5 touches

the parabola y² = 20 x

the ellipse $$9{x^2} + 16{y^2} = 144$$

the hyperbola $${{{x^2}} \over {29}} - {{{y^2}} \over 4} = 1$$

the circle $${x^2} + {y^2} = 25$$

WB JEE 2020

MCQ (More than One Correct Answer)

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Consider a tangent to the ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 1} = 1$$ at any point. The locus of the mid-point of the portion intercepted between the axes is

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

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

$${1 \over {3{x^2}}} + {1 \over {4{y^2}}} = 1$$

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

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