A tangent to the parabola $y^2=8x$ makes an angle ${45}^{\circ }$ with the straight line $y=3x+5$. Find the equation of the tangent and its point of contact.

The equation of any tangent to the parabola 

$y^2=8x$ is $y=mx+\frac{2}{m}$

Clearly,

 $\begin{array}{l}\tan \left({45}^{\circ }\right)=\left|\frac{\mathrm{m}-3}{1+3\mathrm{m}}\right|\\ \Rightarrow \frac{\mathrm{m}-3}{1+3\mathrm{m}}=\pm 1\\ \Rightarrow \frac{\mathrm{m}-3}{1+3\mathrm{m}}=1 and \\ \frac{\mathrm{m}-3}{1+3\mathrm{m}}=-1\end{array}$

  $\Rightarrow \mathrm{m}=-2$, $\frac{1}{2}$

Hence, the equations of tangents are

$\begin{array}{l}\mathrm{y}=-2\mathrm{x}-1, \mathrm{y}=\frac{\mathrm{x}}{2}+4\\ \Rightarrow \mathrm{y}=-2\mathrm{x}-1, \mathrm{x}-2\mathrm{y}+8=0\end{array}$

Solving $y^2=8x$ and $y=-2x-1$ we get, the point of intersection is $\left(\frac{1}{2}, -2\right)$.

Again, solving $y^2=8x$ and $x-2y+8=0$ we get the point of intersection is $(8, 8)$.

Thus, the point of contacts are $\left(\frac{1}{2}, -2\right)$ and $(8, 8)$.