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Let the eccentricity of the hyperbola $\mathrm{\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { be } \frac{5}{4}}$ . If the equation of the normal at the point $\mathrm{\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)}$ on the hyperbola is $\mathrm{8 \sqrt{5} x+\beta y=\lambda,}$ then $\mathrm{ \lambda-\beta}$ is equal to ___________.
Option: 1
85

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

Given

$\mathrm{s: \frac{x^{2}}{a^{2}} -\frac{y^{2}}{b^{2}}=1, e=\frac{5}{4}}\\$

$\mathrm{\because b^{2} =a^{2}\left(e^{2}-1\right)}\\$

$\mathrm{b^{2} =a^{2}\left(\frac{25}{26}-1\right)}\\$

$\mathrm{b^{2} =a^{2}\left(\frac{9}{16}\right)}$ ...........(1)

$\mathrm{a \sec \theta =\frac{8}{\sqrt{5}}} \\$

$\mathrm{\sec ^{2} \theta =\frac{64}{5 a^{2}}}$

$\mathrm{b \tan \theta =\frac{12}{5}}\\$ $......\mathrm{\because b^{2}=a^{2}\left ( \frac{9}{16} \right )}$

$\mathrm{\tan ^{2} \theta =\frac{144}{25 b^{2}}}\\$

$\mathrm{\tan ^{2} \theta =\frac{144}{25} \times \frac{16}{9 a^{2}}}\\$

$\mathrm{\tan ^{2} \theta =\frac{256}{25 a^{2}}}$

$\mathrm{\because \sec ^{2} a-\tan ^{2} a=1 }\\$

$\mathrm{\frac{64}{5 a^{2}}-\frac{25}{25 a^{2}}=1}\\$

$\mathrm{\frac{320-256}{25 a^{2}}=1 \Rightarrow a^{2}=\left(\frac{64}{25}\right)}\\$

$\mathrm{\because b^{2}=a^{2}\left(\frac{9}{16}\right) \Rightarrow b^{2}=\frac{64}{25} \times \frac{9}{16}=\frac{36}{25} }$

The equation of the normal to the hyperola at any point $\mathrm{\left ( x_{1}y_{1} \right ) }$ lying on it is

$\mathrm{\frac{x-x_{1}}{x_{1 / a^{2}}}+\frac{y-y_{1}}{y_{1} / b^{2}}=0}$

$\mathrm{\frac{x-8 / \sqrt{5}}{\frac{8}{\sqrt{5}} / \frac{64}{25}}+\frac{y-12 / 5}{\frac{12}{5}/ \frac{36}{25}}=0}\\$

$\mathrm{\frac{x-\frac{8}{\sqrt{5}}}{\frac{5 \sqrt{5}}{8}}+\frac{y-12 / 5}{5 / 3}=0}$

$\mathrm{\frac{8}{5 \sqrt{5}}\left[x-\frac{8}{\sqrt{5}}\right]+\frac{3}{5}\left[y-\frac{12}{5}\right]=0}\\$

$\mathrm{\frac{8}{5 \sqrt{5}} x-\frac{64}{25}+\frac{3}{5} y-\frac{36}{25}=0}\\$

$\mathrm{\frac{8 \sqrt{5}}{25} x-\frac{64}{25}+\frac{3 y}{5}-\frac{36}{25}=0}$

$\mathrm{\frac{8 \sqrt{5} x+15 y-64-36}{25}=0}\\$

$\mathrm{\text { Ni :} 8 \sqrt{5} x+15 y-100=0}$ ......(1)

$\mathrm{N:8 \sqrt{5} x+\beta y=\lambda }$

$\mathrm{\text { so }{\beta }=15 \quad \lambda=100}\\$

$\mathrm{\lambda-\beta=100-15=85}$

Hence the answer is 85

Posted by

Rakesh

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