A) \[\frac{{{(x-2)}^{2}}}{4}+\frac{{{(y+3)}^{2}}}{3}=1\]
B) \[\frac{{{(x+2)}^{2}}}{3}+\frac{{{(y+3)}^{2}}}{4}=1\]
C) \[\frac{{{(x+2)}^{2}}}{4}+\frac{{{(y+3)}^{2}}}{3}=1\]
D) None of the above
Correct Answer: A
Solution :
Let\[2a\]and\[2b\]be the major and minor axes of the ellipse. Then, its equation is \[\frac{{{(x-2)}^{2}}}{{{a}^{2}}}+\frac{{{(y+3)}^{2}}}{{{b}^{2}}}=1\] ? (i) Here, semi-major axis\[CA=a\] \[\Rightarrow \] \[\sqrt{{{(4-2)}^{2}}+{{(-3+3)}^{2}}}=a\] \[\Rightarrow \] \[a=2\] ...(ii) Here,\[CS=ae\] \[\Rightarrow \] \[\sqrt{{{(2-3)}^{2}}+{{(-3+3)}^{2}}}=ae\] ?(iii) \[ae=1\] ...(iii) From Eqs. (ii) and (iii), we get \[e=\frac{1}{2}\] Now, \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})\] \[\Rightarrow \] \[{{b}^{2}}=4\left( 1-\frac{1}{4} \right)=3\] On substituting the values of a and b in Eq. (i) we get \[\frac{{{(x-2)}^{2}}}{4}+\frac{{{(y+3)}^{2}}}{3}=1\]You need to login to perform this action.
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