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): \(\frac \pi6\).
Top Questions on Hyperbola
- Let the foci of a hyperbola be \( (1, 14) \) and \( (1, -12) \). If it passes through the point \( (1, 6) \), then the length of its latus-rectum is:
- Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $ 2a $ and $ 2b $, respectively, and one focus and the corresponding directrix of this hyperbola be $ (-5, 0) $ and $ 5x + 9 = 0 $, respectively. If the product of the focal distances of a point $ (\alpha, 2\sqrt{5}) $ on the hyperbola is $ p $, then $ 4p $ is equal to:
- If the normal drawn to the hyperbola \( xy = 16 \) at (8, 2) meets the hyperbola again at a point \((\alpha, \beta)\), then \( |\beta| + \frac{1}{|\alpha|} = \)
- If \( 3\sqrt{2}x - 4y = 12 \) is a tangent to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and \(\frac{5}{4}\) is its eccentricity, then \( a^2 - b^2 = \)
- Let H: $\frac{-x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6)$, $\alpha>0$ lies on H. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2 + \beta$ is equal to:
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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.
