यदि रेखा y = mx +7 sqrt{3}, अतिपरवलय frac{x^{ | vedclass

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Question

$\frac{{{x^2}}}{{24}} - \frac{{{y^2}}}{{18}} = 1\,\,\,\,\, \Rightarrow a\, = \sqrt {24} :b\, = \sqrt {18} $

Paramentric normal:

$\sqrt {24} \cos

Vedclass · Accepted Answer

$\frac{2}{{\sqrt 5 }}$

Vedclass · Answer

$\frac{{{x^2}}}{{24}} - \frac{{{y^2}}}{{18}} = 1\,\,\,\,\, \Rightarrow a\, = \sqrt {24} :b\, = \sqrt {18} $

Paramentric normal:

$\sqrt {24} \cos 	heta .x + \sqrt {18} .y\cot 	heta  = 42$

At $x = 0;y = \frac{{42}}{{\sqrt {18} }}	an 	heta  = 7\sqrt 3 $      (from given equation)

$ \Rightarrow 	an 	heta  = \sqrt {\frac{3}{2}}  \Rightarrow \sin 	heta  =  \pm \sqrt {\frac{3}{5}} $

slope of parametric normal $ = \frac{{ - \sqrt {24} \cos 	heta }}{{\sqrt {18} \cot 	heta }} = m$

$ \Rightarrow m =  - \sqrt {\frac{4}{3}} \sin 	heta  =  - \frac{2}{{\sqrt 5 }}$

यदि रेखा $y = mx +7 \sqrt{3}$, अतिपरवलय $\frac{x^{2}}{24}-\frac{y^{2}}{18}=1$ का अभिलंब है, तो $m$ का एक मान है :

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