1.

The line `lx+my+n=0` is a normal to the parabola `y^2 = 4ax` if

Answer» Equation of normal to parabola, `y^2 = 4ax` at point `(at^2,2at)` is given by,
`y+tx = 2at+at^3`
`=tx+y+(-2at-at^3) = 0`
Given equation of the normal to this parabola,
`lx+my+n = 0`
Comparing these two equations,
`l/t = m/1 = n/(-2at-at^3)`
`=>l/t = m/1 => t = l/m`
Now,`m/1 = n/(-2at-at^3)`
`=>m = n/(-2a(l/m)-al^3/m^3)`
`=>m =(nm^3)/(-2alm^2-al^3)`
`=>-2alm^2-al^3 = nm^2`
`=>al(2m^2+l^2)+nm^2 = 0`, which is the required condition.


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