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The Equation of the Directrix of a Hyperbola is X − Y + 3 = 0. Its Focus is (−1, 1) and Eccentricity 3. Find the Equation of the Hyperbola. - Mathematics

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प्रश्न

The equation of the directrix of a hyperbola is x − y + 3 = 0. Its focus is (−1, 1) and eccentricity 3. Find the equation of the hyperbola.

संक्षेप में उत्तर
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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

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

Squaring both the sides, we get:

\[\left( x + 1 \right)^2 + \left( y - 1 \right)^2 = \frac{9}{2} \left( x - y + 3 \right)^2 \]

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

\[ \Rightarrow 2 x^2 + 4x + 2 + 2 y^2 - 4y + 2 = 9 x^2 + 9 y^2 + 81 - 18xy - 54y + 54x\]

\[ \Rightarrow 7 x^2 + 7 y^2 + 50x - 50y - 18xy + 77 = 0\]

Equation of the hyperbola:

\[7 x^2 + 7 y^2 + 50x - 50y - 18xy + 77 = 0\]

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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 1 | पृष्ठ १३

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