Question: If $ x=a,(1+cos\theta ,cos\phi ), $ $ y=b,(1+\cos \theta sin\phi ) $ and $ z=c,(1+\sin \theta ), $ then which one of the following is correct?
Options:
A) $ {{( \frac{x-a}{a} )}^{2}}+{{( \frac{y-b}{b} )}^{2}}+{{( \frac{z-c}{c} )}^{2}}=1 $
B) $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1 $
C) $ x^{2}+y^{2}+z^{2}=a^{2}+b^{2}+c^{2} $
D) $ \frac{{{(x-a)}^{2}}}{a}+\frac{{{(y-b)}^{2}}}{b}+\frac{{{(z-c)}^{2}}}{c}=1 $
Show Answer
Answer:
Correct Answer: A
Solution:
- $ x=a,(1+\cos \theta \cos \phi ) $
$ \Rightarrow $ $ \frac{x}{a}-1=\cos \theta \cos \phi $ … (i)
$ y=b,(1+\cos \theta \sin \phi ) $
$ \Rightarrow $ $ \frac{y}{b}-1=\cos \theta \sin \phi $ … (ii)
and $ z=c,(1+\sin \theta ) $
$ \Rightarrow $ $ \frac{z}{c}-1=\sin \theta $ … (iii)
On squaring and adding Eqs. (i), (ii) and (iii), we get
$ {{( \frac{x-a}{a} )}^{2}}+{{( \frac{y-b}{b} )}^{2}}+{{( \frac{z-c}{c} )}^{2}} $
$ {{\cos }^{2}}\theta ,({{\cos }^{2}}\phi +{{\sin }^{2}}\phi )+{{\sin }^{2}}=1 $
BETA
