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Find the Equation of the Hyperboala Whose Focus is at (5, 2), Vertex at (4, 2) and Centre at (3, 2). - Mathematics

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प्रश्न

Find the equation of the hyperboala whose focus is at (5, 2), vertex at (4, 2) and centre at (3, 2).

थोडक्यात उत्तर
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उत्तर

The equation of the hyperbola with centre (x0,y0) is given by

\[\frac{\left( x - x_0 \right)^2}{a^2} - \frac{\left( y - y_0 \right)^2}{b^2} = 1\]

Focus = \[\left( ae + x_0 , y_0 \right)\]

 Vertex = (a+x0, y0)

\[\therefore ae = 2 \]

and a = 1

\[ b^2 = \left( 2 \right)^2 - a^2 \]

\[ \Rightarrow b^2 = \left( 2^2 \right) - \left( 1 \right)^2 \]

\[ \Rightarrow b^2 = 3\]

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

\[ \Rightarrow 3 \left( x - 3 \right)^2 - \left( y - 2 \right)^2 = 3\]

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पाठ 27: Hyperbola - Exercise 27.1 [पृष्ठ १४]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 27 Hyperbola
Exercise 27.1 | Q 9.1 | पृष्ठ १४
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