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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 (2, 2) directrix is $\text{x+y}=\text{9}$ and eccentricity = 2
✓

Answer

Let S (2, 2) be the focus and P (x, y) be a point on the hyperbola, Draw PM perpendicular from P on the directrix, Then, By definition$\text{sP}=\text{ePM}$
$\Rightarrow\text{sP}^{2}=\text{e}^{2}\text{PM}^{2}$ $\Rightarrow(\text{x}-2)^{2}+(\text{y-2})^{2}=2^{2}\Bigg[\frac{\text{x}+\text{y}-9}{\sqrt{1^{2}+1^{2}}}\Bigg]^{2}$ $\Rightarrow\text{x}^{2}+4-4\text{x}+\text{y}^{2}+4-4\text{y}=\frac{4[\text{x+y}-9]^{2}}{2}$ $\Rightarrow\text{x}^{2}+\text{y}-4\text{x}-\text{4y}+8=2[\text{x+y}-9]^{2}$ $\Rightarrow\text{x}^{2}+\text{y}^{2}-4\text{x}-4\text{y}+8\\=2\Big[\text{x}^{2}+\text{y}^{2}+(-9)^{2}+2\times\text{x}\times\text{y+2}\times\text{y}\times(-9)+2\times(-9)\times\text{x}\Big]$ $\Rightarrow\text{x}^{2}+\text{y}^{2}-4\text{x}-4\text{y}+8=2\Big[\text{x}^{2}+\text{y}^{2}+81+2\text{xy}-18\text{y}+18\text{x}\Big]$ $\Rightarrow\text{x}^{2}+\text{y}^{2}-4\text{x}-4\text{y}+8=\big[2\text{x}^{2}+2\text{y}^{2}+162+4\text{xy}-36\text{y}-36\text{x}$ $\Rightarrow2\text{x}^{2}-\text{x}^{2}+2\text{y}^{2}-\text{y}^{2}+4\text{xy}-36\text{x}+4\text{x}-36\text{y}+4\text{y}+162-8=0$ $\Rightarrow\text{x}^{2}+\text{y}^{2}+4\text{xy}-32\text{x}-32\text{y}+154=0$ This is the required equation of the hyperbola.

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