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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

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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