A) \[{{a}^{2}}{{(CG)}^{2}}+{{b}^{2}}{{(Cg)}^{2}}={{({{a}^{2}}-{{b}^{2}})}^{2}}\]
B) \[{{a}^{2}}{{(CG)}^{2}}-{{b}^{2}}{{(Cg)}^{2}}={{({{a}^{2}}-{{b}^{2}})}^{2}}\]
C) \[{{a}^{2}}{{(CG)}^{2}}-{{b}^{2}}{{(Cg)}^{2}}={{({{a}^{2}}+{{b}^{2}})}^{2}}\]
D) None of these
Correct Answer: A
Solution :
Let at a point \[({{x}_{1}},{{y}_{1}})\] normal will be \[\frac{(x-{{x}_{1}}){{a}^{2}}}{{{x}_{1}}}=\frac{(y-{{y}_{1}}){{b}^{2}}}{{{y}_{1}}}\] At \[G,\,\,y=0\]Þ\[x=CG=\frac{{{x}_{1}}({{a}^{2}}-{{b}^{2}})}{{{a}^{2}}}\] At \[g,\,\,\,x=0\]Þ\[y=Cg=\frac{{{y}_{1}}({{b}^{2}}-{{a}^{2}})}{{{b}^{2}}}\] \[\frac{x_{1}^{2}}{{{a}^{2}}}+\frac{y_{1}^{2}}{{{b}^{2}}}=1\] Þ\[{{a}^{2}}{{(CG)}^{2}}+{{b}^{2}}{{(Cg)}^{2}}={{({{a}^{2}}-{{b}^{2}})}^{2}}.\]You need to login to perform this action.
You will be redirected in
3 sec