Ex 10.4

Chapter 10 Class 11 Conic Sections
Serial order wise

### Transcript

Ex 10.4, 5 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y2 β 9x2 = 36 Given equation is 5y2 β 9x2 = 36. Dividing whole equation by 36 5π¦2/36 β 9π₯2/36 = 36/36 π¦2/((36/5) ) β π₯2/4 = 1 The above equation is of the form π¦2/π2 β π₯2/π2 = 1 β΄ Axis of hyperbola is y-axis Comparing (1) & (2) a2 = 36/5 a = π/βπ & b2 = 4 b = 2 Also, c2 = a2 + b2 c2 = 36/5 + 4 c2 = (36 + 20)/5 c2 = 56/5 c2 = β(56/5) c = (πβππ)/βπ Coβordinate of foci = (0, Β± c) = ("0, Β± " (2β14)/β5) So, co-ordinates of foci are ("0, " (πβππ)/βπ) & ("0," (βπβππ)/βπ) Coordinates of vertices = (0, Β±a) = ("0, " Β±6/β5) So, co-ordinates of vertices are ("0, " π/βπ) & ("0," (βπ)/βπ) Eccentricity is e = π/π e = ((2β14)/β5)/(6/β5) e = (2β14)/β5 Γ β5/6 = βππ/π Latus rectum = 2π2/π = (2 Γ 2^2)/(6/β5) = 2 Γ 4 Γ β5/6 = (πβπ)/π