Let x + y = 0 and x – y = 0 are tangent to a parabola whose focus is S(1, 2). AB be a focal chord of the parabola. If the harmonic mean of AS and BS can be expressed as  $\frac{m}{\sqrt{n}}$ (where m, n are positive integers and m, n are coprime) then the value of (m + 3n) is equal to

Feet of the perpendicular ($N_1$  and  $N_2$) from focus upon any tangent to parabola lies on the tangent line at the vertex.
Now equation of  $S{N}_1$ is x + y = $\lambda$  passing through (1, 2)
 $\Rightarrow \lambda =3$
Equation of  $S{N}_1$  is x + y = 3
Solving x + y = 3 and y = x, we get  $N_1=(\frac{3}{2},\frac{3}{2})$
$\left|\right||ly$  equation of  $S{N}_2$  is x = y = $\mu$  passing through (1, 2)
 $\Rightarrow \mu =-1$
Equation of  $S{N}_2$  is  $y-x=1$
Solving y – x = 1 and y = – x, we get  $N_2\equiv (\frac{-1}{2},\frac{1}{2})$
Now equation of tangent line at vertex is, 2x – 4y + 3 = 0
Distance of S(1, 2) from tangent at vertex is = $\frac{|2-8+3|}{\sqrt{20}}=\frac{3}{2\sqrt{5}}$
=$\frac{1}{4}\times$  latus rectum and hence length of latus rectum =$\frac{6}{\sqrt{5}}$  , harmonic mean of AS and BS is half the length of latus rectum = $\frac{3}{\sqrt{5}}$
Hence m  = 3, n = 5
$\Rightarrow m+3n=18$