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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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Question

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

Answer in Brief

Solution

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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Chapter 27: Hyperbola - Exercise 27.1 [Page 14]

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RD Sharma Mathematics [English] Class 11
Chapter 27 Hyperbola
Exercise 27.1 | Q 9.1 | Page 14

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