Find the equation of ellipse whose focus is $(1, - 2)$ , the directrix $3 x - 2 y + 5 = 0$ and eccentricity equal to $\frac{1}{2}$ .

Open in App

Solution

Given: $e = \frac{1}{2}, S (1, - 2)$ and equation of directrix is $3 x - 2 y + 5 = 0$

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

$\Rightarrow \sqrt{(x - 1)^{2} + (y + 2)^{2}} = \frac{1}{2} \times ∣ ∣ ∣ ∣ ∣ \frac{3 x - 2 y + 5}{\sqrt{3^{2} + (- 2)^{2}}} ∣ ∣ ∣ ∣ ∣$

$\Rightarrow \sqrt{(x - 1)^{2} + (y + 2)^{2}} = \frac{1}{2} \times \frac{| 3 x - 2 y + 5 |}{\sqrt{13}}$

Squaring on both sides, we get

$\Rightarrow (x - 1)^{2} + (y + 2)^{2} = \frac{1}{52} (3 x - 2 y + 5)^{2}$

$\Rightarrow 52 (x^{2} - 2 x + 1 + y^{2} + 4 y + 4)$

$= 9 x^{2} + 4 y^{2} + 25 - 12 x y - 20 y + 30 x$

$\Rightarrow 43 x^{2} + 48 y^{2} + 12 x y - 134 x + 228 y + 235 = 0$

Hence, the equation of ellipse is

$43 x^{2} + 12 x y + 48 y^{2} - 134 x + 228 y + 235 = 0$

Suggest Corrections

Similar questions

(1, - 1)

\frac{2}{3}

x + y + 2 = 0

(1, 0)

x + y + 1 = 0

\frac{1}{\sqrt{2}}

(- 1, 0)

4 x + 3 y + 1 = 0

\frac{1}{\sqrt{5}} .

3 x + 4 y - 5 = 0

(1, 2)

\frac{1}{2}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program