Find the length of perpendicular from the point `(a cos alpha, a sin alpha)` to the line `x cos alpha+y sin alpha=p`.

The length of the perpendicular from the point (a cos alpha, a sin alpha) upon the line y=x tan alpha+c, c le 0, is

If p&q are lengths of perpendicular from the origin x sin alpha+y cos alpha=a sin alpha cos alpha and x cos alpha-y sin alpha=a cos2 alpha, then 4p^(2)+q^(2)

If p & q are lengths of perpendicular from the origin x sin alpha + y cos alpha = a sin alphacos alpha and x cos alpha-y sin alpha = acos 2alpha, then 4p^2 + q^2

The foot of the perpendicular from (0,0) to the line x cos alpha+y sin alpha=p is

The length of the perpendicular from the origin on the line (x sin alpha)/(b ) - (y cos alpha)/(a )-1=0 is

The foot of the perpendicular from (0, 0) to the line x cos alpha+y sin alpha=p is

If p and q are the lengths of the perpendiculars from the origin to the straight lines x "sec" alpha + y " cosec" alpha = a " and " x "cos" alpha-y " sin" alpha = a "cos" 2alpha, " then prove that 4p^(2) + q^(2) = a^(2).