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Hyperbola — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsHyperbola5 Marks
Question
Find the equation of the hyperbola whose,
Focus is (1, 1) directrix is $\text{2x}+\text{y}=1$ and eccentricity $= \sqrt{3}$
✓

Answer

Let S (-1, 1) be the focus and P (x, y) be a point on the hyperbola Draw PM perpendicular from P on the directrix, Then, by definition
$\text{sP}=\text{ePM}$
$\Rightarrow\text{SP}^{2}=\text{e}^{2}\text{PM}^{2}$
$\Rightarrow(\text{x+1})^{2}+(\text{y-1})^{2}=(3)^{2}\Bigg[\frac{\text{x}-\text{y}+3}{\sqrt{1^{2}+(-1)^{2}}}\Bigg]^{2}$
$\Rightarrow\text{x}^{2}+1+2\text{x}+\text{y}^{2}+1-2\text{y}=\frac{9[\text{x}-\text{y}+3]^{2}}{2}$
$\Rightarrow2\text{[x}^{2}+\text{y}^{2}+2\text{x}-2\text{y}+2]=9[\text{x}-\text{y}+3]^{2}$
$\Rightarrow2\text{x}^{2}+2\text{y}^{2}+4\text{x}-4\text{y}+4\\=9\Big[\text{x}^{2}(-\text{y)}^{2}+3^{2}+2\times\text{x}\times\text{x}(-\text{y})\times3+2\times3\times\text{x}\Big]$
$\Rightarrow\text{2}\text{x}^{2}+2\text{y}^{2}+4\text{x}-4\text{y}-4=9\Big[\text{x}^{2}(-\text{y})^{2}+9-2\text{xy}-6\text{y}+\text{6x}$
$\Rightarrow2\text{x}^{2}+2\text{y}^{2}+4\text{x}-4\text{y}+4\\=9\text{x}^{2}+9\text{y}^{2}+81-18\text{xy}-54\text{y}+4\text{y}+81-4=0$
$\Rightarrow7\text{x}^{2}+7\text{y}^{2}-18\text{xy}+50\text{x}-50\text{y}+77=0$
This is the required equation of the hyperbola.

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