Question

Find the equation of the ellipse whose centre is (−2, 3) and whose semi-axis are 3 and 2 when major axis is (i) parallel to x-axis (ii) parallel to y-axis.

Answer 1

\[\text{ When the major axis is parallel to thex-axis. }\]
\[\text{ Let }\frac{(x - x_1 )^2}{a^2} + \frac{(y - y_1 )^2}{b^2} = 1 . . . (1)\]
\[ {\text{ Here, x} }_1 {\text{ and y }}_1 \text{ are } -2 \text { and } 3, \text{ respectively, and 3 and 2 are the lengths of the axes } .\]
\[\text{ Substituting the value in eq. (1), we get }: \]
\[\frac{(x + 2 )^2}{9} + \frac{(y - 3 )^2}{4} = 1\]
\[ \Rightarrow \frac{4( x^2 + 4 + 4x) + 9( y^2 + 9 - 6y)}{36} = 1\]
\[ \Rightarrow 4 x^2 + 16 + 16x + 9 y^2 + 81 - 54y = 36\]
\[ \Rightarrow 4 x^2 + 9 y^2 + 16x - 54y + 61 = 0\]
\[\text{ This is the required equation of the ellipse }.\]
\[(ii) \text{ When the major axis is parallel to they-axis }.\]
\[\text{ Let }\frac{(x - x_1 )^2}{b^2} + \frac{(y - y_1 )^2}{a^2} = 1 . . . (1)\]
\[\text{ Here }, x_1 {\text{ and y }}_1 \text{ are } -2 \text{ and } 3, \text{ respectively, and 3 and 2 are the lengths of the axes }.\]
\[\text{ Substituting the value in eq.(1), we get }: \]
\[\frac{(x + 2 )^2}{4} + \frac{(y - 3 )^2}{9} = 1\]
\[ \Rightarrow \frac{9( x^2 + 4 + 4x) + 4( y^2 + 9 - 6y)}{36} = 1\]
\[ \Rightarrow 9 x^2 + 36 + 36x + 4 y^2 + 36 - 24y = 36\]
\[ \Rightarrow 9 x^2 + 4 y^2 + 36x - 24y + 36 = 0\]
\[\text{ This is the required equation of the ellipse }.\]