Determine the equation of the ellipse whose focus is at $(- 1, 0)$ , directrix is $4 x + 3 y + 1 = 0$ and eccentricity is equal to $\frac{1}{\sqrt{5}} .$

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Solution

Let $S (- 1, 0)$ be the focus and $Z Z^{'}$ be the directrix.
Let $P (x, y)$ be any point on the ellipse and $P M$ be perpendicular from $P$ on the directrix.

Then, by definition
$S P = e . P M$ where $e = \frac{1}{\sqrt{5}}$

On squaring both sides, we get
$S P^{2} = e^{2} P M^{2}$
$\Rightarrow (x + 1)^{2} + (y - 0)^{2} = {(\frac{1}{\sqrt{5}})}^{2} [\frac{4 x + 3 y + 1}{\sqrt{4^{2} + 3^{2}}}]$

$\Rightarrow (x + 1)^{2} + y^{2} = (\frac{1}{5}) [\frac{4 x + 3 y + 1}{\sqrt{25}}]$

$\Rightarrow (x + 1)^{2} + y^{2} = (\frac{1}{25}) [4 x + 3 y + 1]$

$\Rightarrow x^{2} + 1 + 2 x + y^{2} = (\frac{1}{25}) [4 x + 3 y + 1]$

$\Rightarrow 25 x^{2} + 25 + 50 x + 25 y^{2} = 4 x + 3 y + 1$

$\Rightarrow 25 x^{2} + 46 x + 25 y^{2} - 3 y + 24 = 0$

Hence, this is the answer.

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