Let the eccentricity of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) be \(\frac{5}{4}\). If the equation of the normal at the point \((\frac{8}{√5}, \frac{12}{5})\) on the hyperbola is 8√5 x + β y = λ, then λ – β is equal to ____.
Correct Answer: 85
Solution and Explanation
The correct answer is: 85.
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(e=\frac{5}{4})\)
So,
\(b^2=a^2(\frac{25}{16}-1)⇒b=\frac{3}{4}a\)
Also, \((\frac{8}{√5}, \frac{12}{5})\) lies on the given hyperbola.
So, Equation of normal
\(⇒8\sqrt5x+15y=100\)
So,
\(β=15\,and \,λ=100\)
Gives
\(λ-β=85\)
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:
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- 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.
