Ex 11.4, 7

Transcript

Ex 11.4, 7 Find the equation of the hyperbola satisfying the given conditions: Vertices (±2, 0), foci (±3, 0) Vertices are (±2, 0) Hence, vertices are on the x-axis ∴ Equation of hyperbola is of the form ﷐x2﷮a2﷯ – ﷐y2﷮b2﷯ = 1 Now, Co-ordinate of vertices = (±a,0) & Vertices = (±2,0) ∴ (±a, 0) = (±2, 0) Hence a = 2 Also, given coordinates of foci = (±3, 0) And we know that co-ordinates of foci are (±c, 0) ∴ (±c, 0) = (±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 ﷐﷐𝑥﷮2﷯﷮﷐𝑎﷮2﷯﷯ − ﷐﷐𝑦﷮2﷯﷮﷐𝑏﷮2﷯﷯ = 1 ﷐﷐𝑥﷮2﷯﷮﷐2﷮2﷯﷯ − ﷐﷐𝑦﷮2﷯﷮5﷯ = 1 ﷐﷐𝒙﷮𝟐﷯﷮𝟒﷯ − ﷐﷐𝒚﷮𝟐﷯﷮𝟓﷯ = 1