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

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Solution

Let P(x,y) be any point on the ellipse whose focus is S(1,-2) and eccentricity $e = \frac{1}{2} .$
Let PM be perpendicular from P on the directrix. Then,
SP=ePM
$\Rightarrow S P = \frac{1}{2} (P M)$
$\Rightarrow S P^{2} = \frac{1}{4} (P M)^{2}$
$\Rightarrow 4 S P^{2} = (P M)^{2}$
$\Rightarrow 4 [(x - 1)^{2} + (y + 2)^{2}] = {[\frac{3 x - 2 y + 5}{\sqrt{(3)^{2} + (- 2)^{2}}}]}^{2}$
$\Rightarrow 4 [x^{2} + 1 - 2 x + y^{2} + 4 + 4 y] = \frac{(3 x - 2 y + 5)^{2}}{(\sqrt{13})^{2}}$
$\Rightarrow 4 [x^{2} + y^{2} - 2 x + 4 y + 5] = \frac{(3 x - 2 y + 5)^{2}}{13}$
$\Rightarrow 52 [x^{2} + y^{2} - 2 x + 4 y + 5] = (3 x - 2 y + 5)^{2}$
$\Rightarrow 52 x^{2} + 52 y^{2} - 104 x + 208 y + 260 = (3 x - 2 y + 5)^{2}$
$\Rightarrow 52 x^{2} + 52 y^{2} - 104 x + 208 y + 260 = (3 x)^{2} + (- 2 y)^{2} + (5)^{2} + 2 \times 3 x \times (- 2 y) + 2 \times (- 2 y) \times 5 + 2 \times 5 \times 3 x$
$\Rightarrow 52 x^{2} + 52 y^{2} - 104 x + 208 y + 260 = 9 x^{2} + 4 y^{2} + 25 - 12 x y - 20 y + 30 x$
$\Rightarrow 52 x^{2} - 9 x^{2} + 52 y^{2} - 4 y^{2} + 12 x y - 104 x - 30 x + 208 y + 20 y + 260 - 25 = 0$
$\Rightarrow 43 x^{2} + 48 y^{2} + 12 x y - 134 x + 228 y + 235$
=0
This is the required equation of the ellipse.

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