# 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

Find the equation of the ellipse in the case:

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

#### 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