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Ex 11.4, 7 - Find hyperbola: Vertices (2, 0), foci (3, 0)

Ex 11.4,  7 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.4,  7 - Chapter 11 Class 11 Conic Sections - Part 3

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Ex 11.4, 7 Find the equation of the hyperbola satisfying the given conditions: Vertices (Β±2, 0), foci (Β±3, 0) Given Vertices are (Β±2, 0) Hence, vertices are on the x-axis ∴ Equation of hyperbola is of the form π’™πŸ/π’‚πŸ – π’šπŸ/π’ƒπŸ = 1 Now, Co-ordinate of vertices = (Β±a, 0) & Vertices = (Β±2, 0) ∴ (Β±a, 0) = (Β±2, 0) Hence a = 2 Also, Given coordinates of foci = (Β±3, 0) And we know that co-ordinates of foci are (Β±c, 0) ∴ (Β±c, 0) = (Β±3, 0) c = 3 Also c2 = a2 + b2 Putting a = 2, c = 3 32 = 22 + b2 9 = 4 + b2 5 = b2 b2 = 5 Also, Given coordinates of foci = (Β±3, 0) And we know that co-ordinates of foci are (Β±c, 0) ∴ (Β±c, 0) = (Β±3, 0) c = 3 Also c2 = a2 + b2 Putting a = 2, c = 3 32 = 22 + b2 9 = 4 + b2 5 = b2 b2 = 5 Thus, Equation of hyperbola π‘₯^2/π‘Ž^2 βˆ’ 𝑦^2/𝑏^2 = 1 π‘₯^2/2^2 βˆ’ 𝑦^2/5 = 1 𝒙^𝟐/πŸ’ βˆ’ π’š^𝟐/πŸ“ = 1

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