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

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