Saved Bookmarks
| 1. |
Using integration, find the value of `m`. If the area bounded by parabola `y^2=16ax` and the line `y=mx` is `a^2/12` square units. |
|
Answer» `y^2=16an` `y=mx` `(mx)^2=16an` `x=0,x=(16a)/m^2` `int_0^(16/m^2)(sqrt(16ax)-mx)dx` `sqrt(16a)int_0^(16/m^2)x^(1/2)dx-m int_0^((16a)/m^2)*xdx` `sqrt(16a)*2/3*(8*9sqrta)/m^2-m/2*(256a^2)/m^4` `64/3*a^2/m^3-(128a^2)/m^3` `a^2/m^3{(8*64-3*128)/3}` `a^2/m^3*128/3=a^2/12` `m^3=(12*128)/3=4*128` `m=8`. |
|