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ધારો કે અતિવલય $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ ની ઉત્કેન્દ્રતા $\frac{5}{4}$ છે. જો આ અતિવલય પરનાં બિંદુ $\left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ આગળ અભીલંબનું સમીકરણ $8 \sqrt{5} x +\beta y =\lambda$ હોય, તો $\lambda-\beta$ = ............
$89$
$85$
$78$
$45$
Solution
$e ^{2}=1+\frac{ b ^{2}}{ a ^{2}}=\frac{25}{16} \Rightarrow \frac{ b ^{2}}{ a ^{2}}=\frac{9}{16} \ldots \ldots(1)$
$A \left(\frac{8}{\sqrt{5}}, \frac{12}{5}\right)$ 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 $\lambda=100, \beta=15$
$\lambda-\beta=85$
