The point on the line `x-y+ 2= 0` from which the tangent to the parabola `y^2= 8x` is perpendicular to the given line is `(a, b)`, then the line `ax + by+c=0` is

The tangent to the parabola y^(2)=8 x parallel to the line 2 y+x=0 is

The equation of a tangent to the parabola y^(2) =8x is y =x+2 . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

The equation of a tangent to the parabola y^2=""8x"" is ""y""=""x""+""2 . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

The equation of a tangent to the parabola y^(2)=8x is y=x+2. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

The equation of a tangent to the parabola y^2=""8x""i s""y""=""x""+""2 . The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

The equation of a tangent to the parabola y^(2)=12x is 2y+x+12=0. The point on this line such that the other tangent to the parabola is perpendicular to the given tangent is (a,b) then |a+b|=

The equation of the tangent to the parabola y^(2)=16x and perpendicular to the line x-4y-7=0 is

Find the equation of the tangent of the parabola y^(2) = 8x which is perpendicular to the line 2x+ y+1 = 0