Find the locus of the mid-point of the portion of the line `xcosalpha+ysinalpha=p` which is intercepted between the axes.

Answer

N/a

Explanation

Given equation of the line is `xsintheta+ycostheta=p`
Let the mid-point of AB is p(h,k).
So, the mid-point of AB are `(a/(2),(b)/(2))`
Since, the point (a,0) lies on the line (i), then `asintheta+0=p`
`rArrasintheta=prArra= (p)/(sintheta)`
and the point (0,b) also lies on the line then, `0+bcostheta=p`
`rArrbcostheta=prArra=(p).(sintheta)`
Now, mid-point of `AB=(a/(2),(a)/(2))` or `((p)/(2sintheta),(p)/(2costheta))`
`because (p)/(2sintheta)=hrArrsintheta=(p)/(2h)`
`rArr1=(p^2)/(4)((1)/(h^2)+(1)/(k^2))`
Locus of the mid-point is
`4=p^2((1)/(x^2)+(1)/(y^2))`
`rArr 4x^2y^2=p^2(x^2+y^2)`