ધારો કે અતિવલય frac{x^{2}}{a^{2}}-frac{y^{2}} | vedclass

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Question

$e ^{2}=1+\frac{ b ^{2}}{ a ^{2}}=\frac{25}{16} \Rightarrow \frac{ b ^{2}}{ a ^{2}}=\frac{9}{16} \ldots \ldots(1)$

$A \left(\frac{8}{\sqrt{5}}, \fr

Vedclass · Accepted Answer

$85$

Vedclass · Answer

$e ^{2}=1+\frac{ b ^{2}}{ a ^{2}}=\frac{25}{16} \Rightarrow \frac{ b ^{2}}{ a ^{2}}=\frac{9}{16} \ldots \ldots(1)$

$A \left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ satisfies $\frac{ x ^{2}}{ a ^{2}}-\frac{ y ^{2}}{ b ^{2}}=1$

$\Rightarrow \frac{64}{5 a ^{2}}-\frac{144}{25 b ^{2}}=1$

Solving (1) and (2) $b =\frac{6}{5} \quad a =\frac{8}{5}$

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 $\lambda=100, \beta=15$

$\lambda-\beta=85$

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