| 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? |
A) \[{{\left( \frac{x-a}{a} \right)}^{2}}+{{\left( \frac{y-b}{b} \right)}^{2}}+{{\left( \frac{z-c}{c} \right)}^{2}}=1\]
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) \[\frac{{{(x-a)}^{2}}}{a}+\frac{{{(y-b)}^{2}}}{b}+\frac{{{(z-c)}^{2}}}{c}=1\]
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 |
| \[{{\left( \frac{x-a}{a} \right)}^{2}}+{{\left( \frac{y-b}{b} \right)}^{2}}+{{\left( \frac{z-c}{c} \right)}^{2}}\] |
| \[{{\cos }^{2}}\theta \,({{\cos }^{2}}\phi +{{\sin }^{2}}\phi )+{{\sin }^{2}}=1\] |
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