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
\(λ-β=85\)
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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.
