The angle between the tangents drawn from the point ( 3, 4) to the parabola $y^2-2y+4x=0$ , is 

The equation of the parabola is  

$(y-1{)}^2=-4\left(x-\frac{1}{4}\right)$

The equation of any tangent to this parabola is  

$y-1=m\left(x-\frac{1}{4}\right)-\frac{1}{m}$

If it passes through (3, 4), then

$\begin{array}{l}3=\frac{11m}{4}-\frac{1}{m}\\ \Rightarrow 12m=11m^2-4\Rightarrow 11m^2-12m-4=0\end{array}$

Let $m_1, m_2$ be the roots of this equation. Then, 

$m_1+m_2=\frac{12}{11}$and $m_1{m}_2=-\frac{4}{11}$

Let$\theta$ be the angle between the tangents. Then,

$\begin{array}{l}\tan \theta =\frac{m_1-m_2}{1+m_1{m}_2}\\ \Rightarrow \tan \theta =\frac{\sqrt{{{\left(m_1+m_2\right)}^2}_{{ }^{ }}-4m_1{m}_2}}{1+m_1{m}_2}\\ \Rightarrow \tan \theta =\frac{\sqrt{144+176}}{7}=\frac{\sqrt{320}}{7}=\frac{8\sqrt{5}}{7}\\ \Rightarrow \theta ={\tan }^{-1}\left(\frac{8\sqrt{5}}{7}\right)\end{array}$

ALITER The combined equation of the pair of tangents drawn from ( 3, 4) to the parabola 

$y^2-2y+4x=0$ is 

$\begin{array}{l}\left(y^2-2y+4x\right)(16-8+12)=\{4y-(y+4)+2(x+3){\}}^2\\ \Rightarrow 4x^2+12xy-11y^2-72x+12y+4=0\end{array}$

let $\theta$ be the angle between the lines given by this equation. Then, 

$\tan \theta =\left|\frac{2\sqrt{36+44}}{4-11}\right|=\frac{8\sqrt{5}}{7}\left[\because \tan \theta =\frac{2\sqrt{h^2-ab}}{a+b}\right]$