Question

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 ____.

Answer 1

Correct Answer: 85

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

\(λ-β=85\)