question_answer

The, equation of the ellipse whose axes are parallel to the coordinate axes having its centre at the point (2, -3) and focus at (3,-3) and one vertex at (4, -3) is

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                 \[\frac{{{(x-2)}^{2}}}{4}+\frac{{{(y+3)}^{2}}}{3}=1\]