Let the eccentricity of the hyperbola x^2a^2-y^2b^2=1 be 54.

Question

Let the eccentricity of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ be $\dfrac54$. If the equation of the normal at the point $\left(\dfrac8{\sqrt5},\dfrac{12}5\right)$ on the hyperbola is $8\sqrt5x+\beta y=\lambda$, then $\lambda-\beta$ is equal to\_\_\_\_

Answer: 85

Accepted Answer

$e=\dfrac54\Rightarrow b^2=a^2\left(\dfrac{25}{16}-1\right)=\dfrac{9a^2}{16}$.

Substituting the given point into the hyperbola equation and solving with this relation gives $a^2=\dfrac{64}{25},\ b^2=\dfrac{36}{25}$ (i.e. $a=\tfrac85,\ b=\tfrac65$).

Normal: $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$:
$$\frac{64/25}{8/\sqrt5}x+\frac{36/25}{12/5}y=4\ \Longrightarrow\ \frac{8}{5\sqrt5}x+\frac35y=4\ \Longrightarrow\ 8\sqrt5x+15y=100.$$

So $\beta=15,\ \lambda=100$, giving $\lambda-\beta=85.$

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