Question

find the equation of the tangents to the parabola y²= 6xwhich pass through the point (3/2,5). and also find the equation of the tangents to the parabola y²+12x=0 from the point (3,8).

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{parabola}\mathrm{is}, \\ {y}^2=6x......\left(1\right) \\ \mathrm{Comparing}\mathrm{the}\left(1\right)\mathrm{with}{y}^2=4ax,\mathrm{we}\mathrm{get} \\ 4a=6\Rightarrow a=\frac{3}{2} \\ \mathrm{Now},\mathrm{equation}\mathrm{of}\mathrm{tangent}\mathrm{to}\left(1\right)\mathrm{in}\mathrm{slope}\mathrm{form}\mathrm{is}, \\ y=mx+\frac{a}{m} \\ \Rightarrow y=mx+\frac{3}{2m}.......\left(2\right) \\ \mathrm{Since}\mathrm{the}\mathrm{tangent}\mathrm{passes}\mathrm{through}\left(\frac{3}{2},5\right),\mathrm{then}\mathrm{from}\left(2\right),\mathrm{we}\mathrm{get} \\ 5=\frac{3m}{2}+\frac{3}{2m} \\ \Rightarrow 3m^2-10m+3=0 \\ \Rightarrow 3m^2-9m-m+3=0 \\ \Rightarrow 3m\left(m-3\right)-1\left(m-3\right)=0 \\ \Rightarrow \left(m-3\right)\left(3m-1\right)=0 \\ \Rightarrow m-3=0\mathrm{or}3m-1=0 \\ \Rightarrow m=3\mathrm{or}m=\frac{1}{3} \\ \mathrm{When}m=3,\mathrm{then}\mathrm{equation}\mathrm{of}\mathrm{tangent}\mathrm{is}, \\ y=3x+\frac{1}{2}\Rightarrow 6x-2y+1=0 \\ \mathrm{When}m=\frac{1}{3},\mathrm{then}\mathrm{equation}\mathrm{of}\mathrm{tangent}\mathrm{is} \\ y=\frac{x}{3}+\frac{9}{2}\Rightarrow 2x-6y+27=0 \end{gathered}$$