If the line `ax+ by =1` passes through the point of intersection of `y =x tan alpha + p sec alpha, y sin(30^@-alpha)-x cos (30^@-alpha) =p`, and is inclined at `30^@` with `y=tan alpha x` , then prove that `a^2 + b^2 = 3/(4p^2)` .

Explanation

Given lines are
`x "sin" alpha-y "cos" alpha+p=0 " " (1)`
`" and " x "cos" (30^(@)-alpha)-y"sin"(30^(@)-alpha)+p=0`
`" or " x "sin" (60^(@)+alpha)-y"cos"(60^(@)+alpha)+p=0 " " (2)`
Clearly, angle between lines (1) and (2) is `60^(@)`.
Slope of line (1) is tan `alpha` which is parallel to line `y=("tan" alpha)x.`
Given that line ax+by=1 is inclined at `30^(@)` with y `=("tan" alpha)x`.
So, line ax+by+1 is also inclined at `30^(@)` with line (1).
Hence, line ax+by+1 is along acute angle bisector of lines (1) and (2).
Now, acute angle bisector is
`x "sin" alpha-y "cos" alpha+p = -(x "sin"(60^(@)+alpha)-y "cos"(60^(@)+alpha)+p)=0`
`"or " ("sin" alpha+"sin"(60^(@)+ alpha)x-("cos"alpha +"cos"(60^(@) +alpha))y+2p=0`
Comparing this line with ax+by=1, we get
`(a)/("sin"alpha+"sin"(60^(@) + alpha)) = (b)/("cos"alpha+"cos"(60^(@) + alpha)) = (1)/(2p)`
`therefore a=("sin"alpha+"sin"(60^(@) + alpha))/(2p) " and "b = ("cos"alpha+"cos"(60^(@) + alpha))/(2p)`
Squaring and adding, we get
`a^(2) +b^(2) = (1+1+2"cos" 60^(@))/(4p^(2)) rArr a^(2) +b^(2) = (3)/(4p^(3))`