Parabola Part 1

Given, parabola $\Rightarrow y^{2}=6 x$ $\Rightarrow \;\; y^{2}=4\left(\frac{3}{2}\right)x$ and given, line $\Rightarrow 2 x+y=1 .$ $\because$ Equation of any tangent to the parabola having slope $m$ is $y=m x+\frac{3}{2m}$ Slope of line $2 x+y=1$ is $m_{1}=-2$ $\because$ Tangent is perpendicular to this line, $\therefore$ Slope of tangent $=m_{2}=-\;\frac{1}{m}=\;\frac{1}{2}$ $\therefore$ Equation of tangent will be $y=\;\frac{1}{2} x+\;\frac{3}{2} \times 2$ $y=\;\frac{x}{2}+3$ or$\;\; 2 y=x+6$ or$\;\; x-2 y+6=0$ Clearly, on putting the coordinates of point $(5,4)$, the equation of tangent is not satisfied. $\therefore$ Point $(5,4)$ does not lie on this tangent.