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 $\left(\frac{8}{\sqrt{5}},\frac{12}{5}\right)$ on the hyperbola is $8\sqrt{5}x+\beta y=\lambda$, then $\lambda -\beta$ is equal to

Answer 1

Answer: 85

$e^2=1+\frac{b^2}{a^2}=\frac{25}{16}$
$\Rightarrow \frac{b^2}{a^2}=\frac{9}{16}$ ...(1)
$A\left(\frac{8}{\sqrt{5}},\frac{12}{5}\right)$ satisfies $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\Rightarrow \frac{64}{5a^2}-\frac{144}{25b^2}=1$ ...(2)
Solving (1) & (2) $b=\frac{6}{5}a=\frac{8}{5}$
Normal at $A$ is $\frac{\sqrt{5}{a}^{2}x}{8}+\frac{5b^{2}y}{12}=a^2+b^2$
Comparing it $8\sqrt{5}x+\beta y=\lambda$
Gives $\lambda =100,\beta =15$
$\lambda -\beta =85$