If $ x=a \sec \theta \cos \phi, y=b \sec \theta \sin \phi $ and $ z=c \tan \theta $, show that $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1 $


Given:

\( x=a \sec \theta \cos \phi, y=b \sec \theta \sin \phi \) and \( z=c \tan \theta \)

To do:

We have to show that \( \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1 \)

Solution:  

$ x=a \sec \theta \cos \phi$

$\Rightarrow \frac{x}{a}=\sec \theta \cos \phi $
$ y=b \sec \theta \sin \phi$

$\Rightarrow \frac{y}{b}=\sec \theta \sin \phi $
$z=c \tan \theta$

$\Rightarrow \frac{z}{c}=\tan \theta $

Therefore,

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=\sec ^{2} \theta \cos ^{2} \phi+\sec ^{2} \theta \sin ^{2} \phi-\tan ^{2} \theta$
$=\sec ^{2} \theta(\cos ^{2} \phi+\sin ^{2} \phi)-\tan ^{2} \theta$          [Since $\sin ^{2} \theta+\cos ^{2} \theta=1$

$=\sec ^{2} \theta (1)-\tan ^{2} \theta$

$=\sec ^{2} \theta-\tan ^{2} \theta$

$=1$             [Since $\sec ^{2} \theta-\tan ^{2} \theta=1$]

Hence proved.

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Updated on: 10-Oct-2022

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