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Ellipse — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSEllipse4 Marks
Question
Find the equation of the ellips whose focus is (1, -2), the directrix $3\text{x}-2\text{y}+5=0$ and eccentricity equal to $\frac{1}{2}$
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Answer

Let S(1, -2) be the focus and ZZ' be the directrix.
Let P(x, y) be any point on the ellipse and let PM be the perpendicular from P on the directix.
Then by the definition of an ellipse, we have:
$\text{SP}=\text{e}.\text{PM},\text{where e}=\frac{1}{2}$
$\Rightarrow\text{SP}^2=\text{e}^2.\text{PM}^2$
$\Rightarrow(\text{x}-1)^2+(\text{y}+2)^2=\Big(\frac{1}{2}\Big)^2\times\Bigg|\frac{\text{3}\text{x}-2\text{y}+5}{\sqrt{(3)^2+(-2)^2}}\Bigg|^2$
$\Rightarrow\text{x}^2+1-2\text{x}+\text{y}^2+4+4\text{y}\\=\Big(\frac{1}{4}\Big)\times\bigg|\frac{9\text{x}^2+4\text{y}^2+25-12\text{xy}-20\text{y}+30\text{x}}{13}\bigg|$
$\Rightarrow52\big(\text{x}^2+1-2\text{x}+\text{y}^2+4+4\text{y}\big)\\=9\text{x}^2+4\text{y}^2+25-12\text{xy}-20\text{y}+30\text{x}$
$\Rightarrow52\text{x}^2+52-104\text{x}+52\text{y}^2+208+208\text{y}\\=9\text{x}^2+4\text{y}^2+25-12\text{xy}-20\text{y}+30\text{x}$
$\Rightarrow43\text{x}^2+48\text{y}^2-134\text{x}+228\text{y}+235=0$
This is the equation of the required ellipse.

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