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Find the Equation of the Hyperbola Whose Focus is (1, 1) Directrix is 2x + Y = 1 and Eccentricity = √ 3 . - Mathematics

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प्रश्न

Find the equation of the hyperbola whose focus is (1, 1) directrix is 2x + y = 1 and eccentricity = \[\sqrt{3}\].

संक्षेप में उत्तर

उत्तर

 Let be the focus and  \[P\left( x, y \right)\] be any point on the hyperbola.
Draw PM perpendicular to the directrix.
By definition:
SP = ePM

\[\Rightarrow\] \[\sqrt{(x - 1 )^2 + (y - 1 )^2} = \sqrt{3}\left( \frac{2x + y - 1}{\sqrt{5}} \right)\]

Squaring both the sides:

\[(x - 1 )^2 + (y - 1 )^2 = 3 \left( \frac{2x + y - 1}{5} \right)^2 \]

\[ \Rightarrow x^2 + 1 - 2x + y^2 + 1 - 2y = \frac{3}{5}\left( 4 x^2 + y^2 + 1 + 4xy - 2y - 4x \right)\]

\[ \Rightarrow 5 x^2 + 5 - 10x + 5 y^2 + 5 - 10y = 12 x^2 + 3 y^2 + 3 + 12xy - 6y - 12x\]

\[ \Rightarrow 7 x^2 - 2 y^2 + 12xy + 4y - 2x - 7 = 0\]

∴ Equation of the hyperbola = \[7 x^2 - 2 y^2 + 12xy + 4y - 2x - 7 = 0\]

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अध्याय 27: Hyperbola - Exercise 27.1 [पृष्ठ १३]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 27 Hyperbola
Exercise 27.1 | Q 2.3 | पृष्ठ १३

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