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

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प्रश्न

Find the equation of the hyperbola whose focus is (2, −1), directrix is 2x + 3y = 1 and eccentricity = 2 .

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
= ePM

\[\Rightarrow\] \[\sqrt{(x - 2 )^2 + (y + 1 )^2} = 2\left( \frac{2x + 3y - 1}{\sqrt{13}} \right)\] 

Squaring both the sides:

\[(x - 2 )^2 + (y + 1 )^2 = 4 \left( \frac{2x + 3y - 1}{13} \right)^2 \]

\[ \Rightarrow x^2 + 4 - 4x + y^2 + 1 + 2y = \frac{4}{13}\left( 4 x^2 + 9 y^2 + 1 + 12xy - 6y - 4x \right)\]

\[ \Rightarrow 13 x^2 + 52 - 52x + 13 y^2 + 13 + 26y = 16 x^2 + 36 y^2 + 4 + 48xy - 24y - 16x\]

\[ \Rightarrow 3 x^2 + 23 y^2 + 48xy - 50y + 36x - 61 = 0\]

∴ Equation of the hyperbola = \[3 x^2 + 23 y^2 + 48xy - 50y + 36x - 61 = 0\]

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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 2.4 | पृष्ठ १३

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