If the hyperbola \(x² - y² cosec²θ = 5\) and ellipse \(x²cosec²θ + y² = 5\) has eccentricity \(e_H\) and \(e_E\) respectively and \(e_H = \sqrt 7e_E\), then \(θ\) is equal to
\(\frac \pi6\)
\(\frac \pi3\)
\(\frac \pi2\)
\(\frac \pi4\)
The Correct Option is A
Solution and Explanation
The correct option is (A): \(\frac \pi6\).
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Questions Asked in JEE Main exam
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A force \( \vec{f} = x^2 \hat{i} + y \hat{j} + y^2 \hat{k} \) acts on a particle in a plane \( x + y = 10 \). The work done by this force during a displacement from \( (0,0) \) to \( (4m, 2m) \) is Joules (round off to the nearest integer).
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Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
