ChemistryQ1–1 of 1 questions
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1
ChemistryMCQIIT JEE · 1968
ધારો કે $P(a\sec \theta, b\tan \theta)$ અને $Q(a\sec \varphi, b\tan \varphi)$,જ્યાં $\theta + \varphi = \frac{\pi}{2}$,એ અતિવલય $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ પરના બે બિંદુઓ છે. જો $(h, k)$ એ $P$ અને $Q$ આગળના અભિલંબનું છેદબિંદુ હોય,તો $k$ બરાબર શું થાય?
A
$\frac{a^2 + b^2}{a}$
B
$-\left(\frac{a^2 + b^2}{a}\right)$
C
$\frac{a^2 + b^2}{b}$
D
$-\left(\frac{a^2 + b^2}{b}\right)$
Solution
(D) અતિવલય $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ માટે $P(a\sec \theta, b\tan \theta)$ આગળના અભિલંબનું સમીકરણ $ax\sin \theta + by = (a^2 + b^2)\tan \theta$ છે.
તે જ રીતે,$Q(a\sec \varphi, b\tan \varphi)$ આગળના અભિલંબનું સમીકરણ $ax\sin \varphi + by = (a^2 + b^2)\tan \varphi$ છે.
જો $(h, k)$ છેદબિંદુ હોય,તો $y$ માટે ઉકેલતા:
$by(\sin \varphi - \sin \theta) = (a^2 + b^2)(\tan \theta \sin \varphi - \tan \varphi \sin \theta)$.
$\varphi = \frac{\pi}{2} - \theta$ હોવાથી,$\sin \varphi = \cos \theta$ અને $\tan \varphi = \cot \theta$ મળે.
$by(\cos \theta - \sin \theta) = (a^2 + b^2)(\sin \theta - \cos \theta)$.
તેથી,$by = -(a^2 + b^2)$,એટલે કે $k = -\frac{a^2 + b^2}{b}$.
તે જ રીતે,$Q(a\sec \varphi, b\tan \varphi)$ આગળના અભિલંબનું સમીકરણ $ax\sin \varphi + by = (a^2 + b^2)\tan \varphi$ છે.
જો $(h, k)$ છેદબિંદુ હોય,તો $y$ માટે ઉકેલતા:
$by(\sin \varphi - \sin \theta) = (a^2 + b^2)(\tan \theta \sin \varphi - \tan \varphi \sin \theta)$.
$\varphi = \frac{\pi}{2} - \theta$ હોવાથી,$\sin \varphi = \cos \theta$ અને $\tan \varphi = \cot \theta$ મળે.
$by(\cos \theta - \sin \theta) = (a^2 + b^2)(\sin \theta - \cos \theta)$.
તેથી,$by = -(a^2 + b^2)$,એટલે કે $k = -\frac{a^2 + b^2}{b}$.
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