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

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

$λ-β=85$