In the curve `x^(m+n)=a^(m-n)y^(2n)` , prove that the `m t h` power of the sub-tangent varies as the `n t h` power of the sub-normal.

Options

A. `m`
B. `n`
C. `1//n`
D. `1//m`

Answer

B

Explanation

Given `x^(m+n)=a^(m-n)y^(2n)`
On taking log both sides, we get
`(m+n)logx(m-n)log a +2n log y`
On differentiating both sides w.r.t `x` we get
`((m+n))/x=0+(2n)/y(dy)/(dx)implies(dy)/(dx)=((m+n))/(2n)(y/x)`
Now, `(("subtangent")^(m))/(("subnormal")^(n))=((y(dx)/(dy))^(m))/((y(dy)/(dx))^(n))=((y^(m-n))/((dy)/(dx))^(m+n))`
`=(x^(m+n))/(((m+n)/(2n))^(m+n)y^(2n))`
[ On putting the value of `(dy)/(dx)`]
`=((2n)/(m+n))^(m+n)xxa^(m-n) [ :' (x^(m+n))/(y^(2n))=a^(m-n)]`
`=` Constant it is [ `:.` independent of `x` and `y` ]
`implies("Subtangent")^(m)prop("Subnormal")^(n)`