Let the eccentricity of the hyperbola x 2 a 2 - y 2 b 2 = 1 be 5 4 . If the equation of the normal at the point 8 5 , 12 5 on the hyperbola is 8 5 x + β y = λ , then λ - β is equal to _____.

Question

Let the eccentricity of the hyperbola x 2 a 2 - y 2 b 2 = 1 be 5 4 . If the equation of the normal at the point 8 5 , 12 5 on the hyperbola is 8 5 x + &#946; y = &#955; , then &#955; - &#946; is equal to _____.

Aicharya · Accepted Answer

Using the formula of eccentricity we get,

$e^{2} = 1 + \frac{b^{2}}{a^{2}} = \frac{25}{16} \Rightarrow \frac{b^{2}}{a^{2}} = \frac{9}{16} . . . (1)$

$A (\frac{8}{\sqrt{5}}, \frac{12}{5})$ 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 $8 \sqrt{5} x + β y = λ$

Gives $λ = 100, β = 15$

So, $λ - β = 85$

Let the eccentricity of the hyperbola x 2 a 2 - y 2 b 2 = 1 be 5 4 . If the equation of the normal at the…

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