JEE Main 2022 (Online) 25th June Evening ShiftQuestion 1
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 $\_\_\_\_\_\_$ .
Hyperbola
Answer$85$
Solution
$e=\tfrac54\Rightarrow b^2=\tfrac{9}{16}a^2$. Point on the curve: $\dfrac{64}{5a^2}-\dfrac{144}{25b^2}=1\Rightarrow\dfrac{64}{5a^2}-\dfrac{256}{25a^2}=1\Rightarrow a^2=\dfrac{64}{25}$, $b^2=\dfrac{36}{25}$. Normal: $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2\Rightarrow\dfrac{8\sqrt5}{25}x+\dfrac35y=4\Rightarrow 8\sqrt5x+15y=100$. $\beta=15$, $\lambda=100$, so $\lambda-\beta=85$.