A) \[Y=\frac{1}{2}\times 3\times {{10}^{10}}\times \frac{1}{4\times {{10}^{4}}}=\frac{3}{8}\times {{10}^{6}}\]
B) \[=0.375\times {{10}^{6}}\]
C) \[=3.75\times {{10}^{5}}\]
D) None of the above
Correct Answer: C
Solution :
Given equation of second ellipse can be rewritten as \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{1}=1\] Equation of tangent to the ellipse is \[\frac{x}{2}\cos \theta +y\sin \theta =1\] ...(i) Equation of the first ellipse can be rewritten as \[\frac{{{x}^{2}}}{6}+\frac{{{y}^{2}}}{3}=1\] ??.. (ii) Let ellipse (i) meets the first ellipse at P and Q and the tangents at P and Q to the second ellipse intersected at \[(h,k)\], then Eq, (i) is the chord of contact of (A, k) with respect to the ellipse (ii). Thus. its equation is \[\frac{hx}{6}+\frac{ky}{3}=1\] ?? (ii) Since, Eqs. (i) and (iii) represent the same line. \[\frac{h/6}{\cos \frac{\theta }{2}}=\frac{k/3}{\sin \theta }=1\] Now, \[{{\text{h}}^{\text{2}}}+{{\text{k}}^{\text{2}}}={{\text{3}}^{\text{2}}}\text{co}{{\text{s}}^{\text{2}}}\theta +\text{ }{{\text{3}}^{\text{2}}}\text{si}{{\text{n}}^{\text{2}}}\theta \] \[=\text{9}\left( \text{1} \right)=\text{9}\] Hence, locus is \[{{x}^{2}}+{{y}^{2}}=9\].
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