Find the equation of ellipse when: Focus is $(- 1, 1)$ , directrix is $x - y + 3 = 0$ and $e = \frac{1}{2}$

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Solution

Given: $e = \frac{1}{2}, S (- 1, 1)$ and equation of directrix is $x - y + 3 = 0$

Let a point $P (x, y)$ , such that

$S P = e \cdot P M$ , where $P M$ is perpendicular distance from $P (x, y)$ to directrix

$\Rightarrow \sqrt{(x + 1)^{2} + (y - 1)^{2}} = \frac{1}{2} \times ∣ ∣ ∣ ∣ ∣ \frac{x - y + 3}{\sqrt{1^{2} + (- 1)^{2}}} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow \sqrt{(x + 1)^{2} + (y - 1)^{2}} = \frac{1}{2} \times \frac{| x - y + 3 |}{\sqrt{2}}$

Squaring on both sides, we get

$\Rightarrow (x + 1)^{2} + (y - 1)^{2} = \frac{1}{8} (x - y + 3)^{2}$

$\Rightarrow 8 (x^{2} + 2 x + 1 + y^{2} - 2 y + 1)$

$= x^{2} + y^{2} + 9 - 2 x y - 6 y + 6 x$

$\Rightarrow 7 x^{2} + 7 y^{2} + 2 x y + 10 x - 10 y + 7 = 0$

Hence, the equation of ellipse is

$7 x^{2} + 2 x y + 7 y^{2} + 10 x - 10 y + 7 = 0$

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Similar questions

Find the equation of the ellipse in the following cases:
(i) focus is (0,1), directrix is x+y=0 and $e = \frac{1}{2}$ .
(ii) focus is (-1,1), directrix is x-y+3=0 and $e = \frac{1}{2}$ .
(iii) focus is (-2,3), directrix is 2x+3y+4=0 and $e = \frac{4}{5}$ .
(iv) focus is (1,2), directrix is 3x+4y-5=0 and $e = \frac{1}{2}$ .