અતિવલય frac{{{x^2}}}{4} - frac{{{y^2}}}{2} = | vedclass

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Question

equation of the tangent at the point $'	heta '$ is

$\frac{{x\sec 	heta }}{a} - \frac{{y	an 	heta }}{b} = 1$

$ \Rightarrow P = \

Vedclass · Accepted Answer

$\frac{4}{{{x^2}}} - \frac{2}{{{y^2}}} = 1$

Vedclass · Answer

equation of the tangent at the point $'	heta '$ is

$\frac{{x\sec 	heta }}{a} - \frac{{y	an 	heta }}{b} = 1$

$ \Rightarrow P = \left( {a\cos 	heta ,0} ight)\,\,\,\,Q = \left( {0, - b\cot 	heta } ight)$ 

Let $R$ be $\left( {h,k} ight) \Rightarrow h = a\cos 	heta ,k =  - b\cot 	heta $

$ \Rightarrow \frac{k}{h} = \frac{{ - b}}{{a\sin 	heta }} \Rightarrow \sin 	heta  = \frac{{ - bh}}{{ak}}$

$\cos 	heta  = \frac{h}{a}$

By squaring and adding,

$\frac{{{b^2}{h^2}}}{{{a^2}{k^2}}} + \frac{{{h^2}}}{{{a^2}}} = 1$

$ \Rightarrow \frac{{{b^2}}}{{{k^2}}} + 1 = \frac{{{a^2}}}{{{h^2}}}$

$ \Rightarrow \frac{{{a^2}}}{{{h^2}}} - \frac{{{b^2}}}{{{k^2}}} = 1$

Now, given $e{q^n}$ of hyperbola is $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{2} = 1$

$ \Rightarrow {a^2} = 4,{b^2} = 2$

$	herefore $ $R$ lies on $\frac{{{a^2}}}{{{x^2}}} - \frac{{{b^2}}}{{{y^2}}} = 1$ i.e., $\frac{4}{{{x^2}}} - \frac{2}{{{y^2}}} = 1$

Similar Questions

રેખા $y = \alpha x + \beta $ એ અતિવલય $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ નો સ્પર્શક હોય તો ચલિતબિંદુ $P(\alpha ,\,\beta )$ નો બિંદુગણ મેળવો.

બિંદુ $\left( {a\,\,\sec \,\theta ,\,\,b\,\,\tan \,\,\theta } \right)$ આગળ અતિવલય $\frac{{{x^2}}}{{{a^2}}}\,\, - \,\,\frac{{{y^2}}}{{{b^2}}}\,\, = \,\,1\,$ ના અભિલંબનું સમીકરણ મેળવો.

અતિવલય $16x^2 - 9y^2 = 14$ નો નાભિલંબની લંબાઈ મેળવો.

આપેલ શરતોનું પાલન કરતાં અતિવલયનું સમીકરણ મેળવો : નાભિઓ $(\pm 5,\,0),$ મુખ્ય અક્ષની લંબાઈ $8$