A) \[2{{a}^{2}}\]
B) \[2{{b}^{2}}\]
C) \[{{a}^{2}}+{{b}^{2}}\]
D) \[{{a}^{2}}-{{b}^{2}}\]
Correct Answer: A
Solution :
[a] \[\sqrt{{{a}^{2}}-{{b}^{2}}}=\pm ae\,\,(0,\,\,ae)\] So, the points are (ae, 0) and (0, -ae). Let \[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\] be a tangent then sum of squares of perpendicular from these points is \[=\frac{{{\left( 1-\frac{ae}{b}\sin \theta \right)}^{2}}+{{\left( 1+\frac{ae}{b}\sin \theta \right)}^{2}}}{\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}}\] \[=\frac{2\left( 1+\frac{{{a}^{2}}{{e}^{2}}}{{{b}^{2}}}{{\sin }^{2}}\theta \right)}{\frac{{{\cos }^{2}}\theta }{{{a}^{2}}}+\frac{{{\sin }^{2}}\theta }{{{b}^{2}}}}\] \[=2{{a}^{2}}\left( \frac{{{b}^{2}}+({{a}^{2}}-{{b}^{2}})si{{n}^{2}}\theta }{{{b}^{2}}(1-sin\theta )+{{a}^{2}}{{\sin }^{2}}\theta } \right)=2{{a}^{2}}\]
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