Find the equation of ellipse when: Focus is $(- 2, 3)$ , directrix is $2 x + 3 y + 4 = 0$ and $e = \frac{4}{5}$

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Solution

Given: $e = \frac{4}{5}, S (- 2, 3)$ and equation of directrix is
$2 x + 3 y + 4 = 0$

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

$\Rightarrow \sqrt{(x + 2)^{2} + (y - 3)^{2}} = \frac{4}{5} \times ∣ ∣ ∣ ∣ \frac{2 x + 3 y + 4}{\sqrt{2^{2} + 3^{2}}} ∣ ∣ ∣ ∣$

$\Rightarrow \sqrt{(x + 2)^{2} + (y - 3)^{2}} = \frac{4}{5} \times \frac{| 2 x + 3 y + 4 |}{\sqrt{13}}$

Squaring on both sides, we get

$\Rightarrow (x + 2)^{2} + (y - 3)^{2} = \frac{16}{25 \times 13} (2 x + 3 y + 4)^{2}$

$\Rightarrow 325 (x^{2} + y^{2} + 4 x - 6 y + 13)$

$= 16 (2 x + 3 y + 4)^{2}$

Hence, the equation of ellipse is

$325 (x^{2} + y^{2} + 4 x - 6 y + 13) = 16 (2 x + 3 y + 4)^{2}$

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Similar questions

Q. 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}

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 $e = \frac{1}{2}$ .