Let the eccentricity of the hyperbola
\(H : \frac{x²}{a²} - \frac{y²}{b²} = 1\)
be √(5/2) and length of its latus rectum be 6√2, If y = 2x + c is a tangent to the hyperbola H. then the value of c2 is equal to
Let the eccentricity of the hyperbola
\(H : \frac{x²}{a²} - \frac{y²}{b²} = 1\)
be √(5/2) and length of its latus rectum be 6√2, If y = 2x + c is a tangent to the hyperbola H. then the value of c2 is equal to
18
20
24
32
The Correct Option is B
Solution and Explanation
The correct answer is (B) : 20
\(1 + \frac{b²}{a²} = \frac{5}{2} ⇒ \frac{b²}{a²} = \frac{3}{2}\)
\(\frac{2b²}{a} = 6\sqrt2 = 2 . \frac{3}{2} . a = 6\sqrt2\)
\(⇒ a = 2\sqrt2 , b² = 12\)
\(c² = a²m² - b² = 8.4 - 12 = 20\)
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.
