If the line x – 1 = 0 is a directrix of the hyperbola kx2 – y2 = 6, then the hyperbola passes through the point
If the line x – 1 = 0 is a directrix of the hyperbola kx2 – y2 = 6, then the hyperbola passes through the point
\((-2\sqrt5,6)\)
\((-\sqrt5,3)\)
\((\sqrt5,-2)\)
\((2\sqrt5,3\sqrt6)\)
The Correct Option is C
Solution and Explanation
Given hyperbola : \(\frac{x^2}{6/k}-\frac{y^2}{6} = 1\)
\(e = \sqrt{1+\frac{6}{6/k}}=\sqrt{1+k}\)
\(x = ± \frac{a}{e} ⇒ x = ± \frac{\sqrt6}{\sqrt{k}\sqrt{k+1}}\)
As given : \(\frac{\sqrt6}{\sqrt{k}\sqrt{k+1}}=1\)
\(⇒ k = 2\)
\(⇒ \frac{x^2}{3}-\frac{y^2}{6} = 1\)
Hence, the option that satisfies and is the correct option is (C):\( (\sqrt{5},-2)\)
Top Questions on Hyperbola
- If the line \( ax + 2y = 1 \), where \( a \in \mathbb{R} \), does not meet the hyperbola \( x^2 - 9y^2 = 9 \), then a possible value of \( a \) is:
- Let \( PQ \) be a chord of the hyperbola \[ \frac{x^2}{4} - \frac{y^2}{b^2} = 1, \] perpendicular to the \(x\)-axis such that \( OPQ \) is an equilateral triangle, \( O \) being the centre of the hyperbola. If the eccentricity of the hyperbola is \( \sqrt{3} \), then find the area of the triangle \( OPQ \).
- 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 \( |\beta| + \frac{1}{|\alpha|} = \)
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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.
