A) \[{{b}^{2}}(1-{{e}^{2}})(2-{{e}^{2}})\]
B) \[{{a}^{2}}({{e}^{4}}-{{e}^{2}}+2)\]
C) \[{{a}^{2}}(1+{{e}^{2}})(2+{{e}^{2}})\]
D) \[{{b}^{2}}(1+{{e}^{2}})(2+{{e}^{2}})\]
Correct Answer: B
Solution :
[b] Normal at \[P(\theta )\]is \[\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }={{a}^{2}}-{{b}^{2}}\] ...(i) Normal at \[P\left( \frac{\pi }{2}+\theta \right)\]is \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] Or \[-\frac{ax}{\sin \theta }-\frac{by}{\cos \theta }={{a}^{2}}-{{b}^{2}}\] ...(ii) Equation (i) and (ii) meet the major axis at \[G\left( \frac{({{a}^{2}}-{{b}^{2}})cos\theta }{a},0 \right)\]and \[g\left( \frac{({{a}^{2}}-{{b}^{2}})sin\theta }{a},0 \right)\] Now,| \[P{{G}^{2}}+Q{{g}^{2}}={{\left\{ \frac{({{a}^{2}}-{{b}^{2}})cos\theta }{a}-a\cos \theta \right\}}^{2}}+{{(0-bsin\theta )}^{2}}\] |
| \[+{{\left\{ \frac{({{a}^{2}}-{{b}^{2}})cos\theta }{a}-a\cos \theta \right\}}^{2}}+{{(0-bcos\theta )}^{2}}\] |
| \[=\frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{a}^{2}}}+{{b}^{2}}+{{a}^{2}}\] |
| \[={{a}^{2}}\left\{ \frac{{{({{a}^{2}}-{{b}^{2}})}^{2}}}{{{a}^{4}}}+\frac{{{b}^{2}}}{{{a}^{2}}}+1 \right\}\] |
| \[={{a}^{2}}\left\{ {{\left( 1-\frac{{{b}^{2}}}{{{a}^{2}}} \right)}^{2}}+\frac{{{b}^{2}}}{{{a}^{2}}}+1 \right\}\] |
| \[={{a}^{2}}({{e}^{4}}+2-{{e}^{2}})\] |
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