A tangent to the parabola $${y^2} = 8x,$$   which makes an angle of $${45^ \circ }$$  with the straight line $$y = 3x + 5$$   is :

Solution :
We know the tangent to the parabola $${y^2} = 4ax$$   at $$\left( {a{t^2},\,2at} \right)$$   is $$ty = x + a{t^2}.$$
Here $$a = 2$$
So, the tangent at $$\left( {2{t^2},\,4t} \right)$$   to the parabola
$${y^2} = 8x$$   is $$ty = x + 2{t^2}......\left( {\text{i}} \right)$$
$$'m'$$ of $$\left( {\text{i}} \right)$$ is $$\frac{1}{t};\left( {\text{i}} \right)$$  makes $${45^ \circ }$$  with $$y = 3x + 5$$   if
$$\eqalign{ & \tan \,{45^ \circ } = \left| {\frac{{\frac{1}{t} - 3}}{{1 + \frac{1}{t}.3}}} \right| = \left| {\frac{{1 - 3t}}{{t + 3}}} \right| \cr & \therefore \,1 = \left| {\frac{{1 - 3t}}{{t + 3}}} \right|{\text{; or }}\frac{{1 - 3t}}{{t + 3}} = \pm 1\,;{\text{ or }}1 - 3t = t + 3,\, - t - 3 \cr & \therefore \,4t = - 2{\text{ or }}2t = 4 \cr & \therefore \,t = - \frac{1}{2}{\text{ or }}2 \cr} $$
Putting in $$\left( {\text{i}} \right),$$ the tangents have the equations
$$\eqalign{ & - \frac{1}{2}y = x + 2.\frac{1}{4}{\text{ i}}{\text{.e}}{\text{., }}2x + y + 1 = 0 \cr & {\text{and }}2y = x + 2.4{\text{ i}}{\text{.e}}{\text{., }}x - 2y + 8 = 0 \cr} $$