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Find the Equation of the Ellipse in the Case: (Iii) Focus is (−2, 3), Directrix is 2x + 3y + 4 = 0 and E = 4 5 - Mathematics

Answer in Brief

Find the equation of the ellipse in the case: 

focus is (−2, 3), directrix is 2x + 3y + 4 = 0 and e = \[\frac{4}{5}\]

 
 

 

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Solution

\[\text{ Let S( - 2, 3) be the focus and ZZ' be the directrix . } \]
\[\text{ Let P(x, y) be any point on the ellipse and let PM be the perpendicular from P on the directrix }  . \]
\[\text{ Then by the definition, we have: } \]
\[SP = e \times PM\]
\[ \Rightarrow SP = \frac{4}{5} \times PM\]
\[ \Rightarrow \frac{5}{4}SP = PM\]
\[ \Rightarrow \frac{25}{16} \left( SP \right)^2 = {PM}^2 \]
\[ \Rightarrow \frac{25}{16}\left[ \left( x + 2 \right)^2 + \left( y - 3 \right)^2 \right] = \left| \frac{2x + 3y + 4}{\sqrt{2^2 + 3^2}} \right|^2 \]
\[ \Rightarrow \frac{25}{16}\left[ x^2 + 4 + 4x + y^2 + 9 - 6y \right] = \frac{4 x^2 + 9 y^2 + 16 + 12xy + 24y + 16x}{13}\]
\[ \Rightarrow 325\left( x^2 + 4 + 4x + y^2 + 9 - 6y \right) = 16\left( 4 x^2 + 9 y^2 + 16 + 12xy + 24y + 16x \right)\]
\[ \Rightarrow 325 x^2 + 1300 + 1300x + 325 y^2 + 2925 - 1950y = 64 x^2 + 144 y^2 + 256 + 192xy + 384y + 256x\]
\[ \Rightarrow 261 x^2 + 181 y^2 + 1044x - 2309y - 192xy + 3969 = 0\]
\[\text{ This is the required equation of the ellipse } .\]

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Exercise 26.1 | Q 2.3 | Page 22
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