The equation of a conic with directrix $3 x + 4 y - 5 = 0$ and focus (0, 0) is given as $x^{2} + y^{2} = (3 x + 4 y - 5)^{2}$ . Find eccentricity of the conic.
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Solution

For a conic, the ratio of distance from focus and perpendicular distance from directrix is equal to eccentricity. We will use this relation to find the eccentricity.

Let (x, y) be the point on the conic.

$P S = \sqrt{x^{2} + y^{2}}$

$P M = ∣ ∣ ∣ \frac{3 x + 4 y - 5}{\sqrt{3^{2} + 4^{2}}} ∣ ∣ ∣ = ∣ ∣ \frac{3 x + 4 y - 5}{5} ∣ ∣$

$\frac{P S}{P M} = e \Rightarrow P S = e P M$

$\Rightarrow P S^{2} = e^{2} P M^{2}$

$\Rightarrow x^{2} + y^{2} = e^{2} \frac{(3 x + 4 y - 5)^{2}}{25}$

Comparing it with given equation ,we get

$\frac{e^{2}}{25} = 1$

$\Rightarrow e = \mp 5$

$\Rightarrow e = 5$ (e is ratio of distances)

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The equation of a conic with directrix 3x + 4y − 5 = 0 and focus (0, 0) is given as x2 + y2 = (3x + 4y − 5)2. Find eccentricity of the conic. __

The equation of a conic with directrix $3 x + 4 y - 5 = 0$ and focus (0, 0) is given as $x^{2} + y^{2} = (3 x + 4 y - 5)^{2}$ . Find eccentricity of the conic.
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