Ex 11.4, 5

Transcript

Ex 11.4, 5 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y2 – 9x2 = 36 Given equation is 5y2 – 9x2 = 36. Dividing whole equation by 36 ﷐5𝑦2﷮36﷯ − ﷐9𝑥2﷮36﷯ = ﷐36﷮36﷯ ﷐𝑦2﷮﷐﷐36﷮5﷯﷯﷯ − ﷐𝑥2﷮4﷯ = 1 The above equation is of the form ﷐𝑦2﷮𝑎2﷯ − ﷐𝑥2﷮𝑏2﷯ = 1 ∴ Axis of hyperbola is y-axis Comparing (1) & (2) a2 = ﷐36﷮5﷯ a = ﷐𝟔﷮﷐﷮𝟓﷯﷯ & b2 = 4 b = 2 Also, c2 = a2 + b2 c2 = ﷐36﷮5﷯ + 4 c2 = ﷐36 + 20﷮5﷯ c2 = ﷐56﷮5﷯ c2 = ﷐﷮﷐56﷮5﷯﷯ c = ﷐𝟐﷐﷮𝟏𝟒﷯﷮﷐﷮𝟓﷯﷯ Co−ordinate of foci = (0, ± c) = ﷐0, ± ﷐2﷐﷮14﷯﷮5﷯﷯ So, co-ordinates of foci are ﷐0, ﷐2﷐﷮14﷯﷮5﷯﷯ & ﷐0, −﷐2﷐﷮14﷯﷮5﷯﷯ Coordinates of vertices = (0, ±a) = (0, ±﷐6﷮﷐﷮5﷯﷯) So, co-ordinates of vertices are ﷐0, ﷐6﷮﷐﷮5﷯﷯﷯ & ﷐0,﷐−6﷮﷐﷮5﷯﷯﷯ Eccentricity is e = ﷐𝑐﷮𝑎﷯ e = ﷐﷐2﷐﷮14﷯﷮﷐﷮5﷯﷯﷮﷐6﷮﷐﷮5﷯﷯﷯ e = ﷐2﷐﷮14﷯﷮﷐﷮5﷯﷯ × ﷐﷐﷮5﷯﷮6﷯ = ﷐﷐﷮14﷯﷮3﷯ Latus rectum = ﷐2𝑏2﷮𝑎﷯ = ﷐2 × ﷐2﷮2﷯﷮﷐6﷮﷐﷮5﷯﷯﷯ = 2 × 4 × ﷐﷐﷮5﷯﷮6﷯ = ﷐4﷐﷮5﷯﷮3﷯