Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Find the Equation of the Ellipse Whose Focus is (1, −2), the Directrix 3x − 2y + 5 = 0 and Eccentricity Equal to 1/2. - Mathematics

Find the equation of the ellipse whose focus is (1, −2), the directrix 3x − 2y + 5 = 0 and eccentricity equal to 1/2.

#### Solution

$\text{ Let S(1, - 2) be the focus and ZZ' be the directrix } .$
$\text{ Let P(x, y) be any point on the ellipse and let PM be the perpendicular from P on the directix } .$
$\text{ Then by the definition of an ellipse, we have:}$
$SP = e .\text{ PM, where e } = \frac{1}{2}$
$\Rightarrow S P^2 = e^2 . P M^2$
$\Rightarrow (x - 1 )^2 + (y + 2 )^2 = \left( \frac{1}{2} \right)^2 . \left| \frac{3x - 2y + 5}{\sqrt{(3 )^2 + ( - 2 )^2}} \right|^2$
$\Rightarrow x^2 + 1 - 2x + y^2 + 4 + 4y = \left( \frac{1}{4} \right) . \left| \frac{9 x^2 + 4 y^2 + 25 - 12xy - 20y + 30x}{13} \right|$
$\Rightarrow 52( x^2 + 1 - 2x + y^2 + 4 + 4y) = 9 x^2 + 4 y^2 + 25 - 12xy - 20y + 30x$
$\Rightarrow 52 x^2 + 52 - 104x + 52 y^2 + 208 + 208y = 9 x^2 + 4 y^2 + 25 - 12xy - 20y + 30x$
$\Rightarrow 43 x^2 + 48 y^2 - 134x + 228y + 12xy + 235 = 0$
$\text{ This is the equation of the required ellipse } .$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 26 Ellipse
Exercise 26.1 | Q 1 | Page 22