the equation of a tangent to the parabola y²=8x is y=x+2. prove that The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is (-2,0).

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Given}, \\ {\mathrm{y}}^2=8\mathrm{x} \\ \mathrm{Equation} \mathrm{of} \mathrm{the} \mathrm{tangent} \mathrm{is} \mathrm{y}=\mathrm{x}+2 \\ \mathrm{Slope} \mathrm{of} \mathrm{the} \mathrm{tangent} \mathrm{is}, \mathrm{m}=1 \\ \mathrm{When} 2 \mathrm{lines} \mathrm{are} \mathrm{perpendicular} \mathrm{to} \mathrm{each} \mathrm{other}, \mathrm{then} \\ \mathrm{product} \mathrm{of} \mathrm{their} \mathrm{slopes} \mathrm{is} -1. \\ \mathrm{m}.{\mathrm{m}}_1=-1 \\ {\mathrm{m}}_1=\frac{-1}{1}=-1 \\ \mathrm{let} \mathrm{other} \mathrm{tangent} \mathrm{be}, \\ \mathrm{y}={\mathrm{m}}_1\mathrm{x}+\mathrm{c}=-\mathrm{x}+\mathrm{c} \\ \mathrm{A} \mathrm{line} \mathrm{touching} \mathrm{the} \mathrm{parabola} \mathrm{is} \mathrm{said} \mathrm{to} \mathrm{be} \mathrm{a} \mathrm{tangent} \mathrm{to} \mathrm{the} \mathrm{parabola} \mathrm{provided} \mathrm{it} \mathrm{satisfies} \mathrm{certain} \mathrm{conditions}. \\ \mathrm{If} \mathrm{we} \mathrm{have} \mathrm{a} \mathrm{line} \mathrm{y} = {\mathrm{m}}_1\mathrm{x} + \mathrm{c} \mathrm{touching} \mathrm{a} \mathrm{parabola} {\mathrm{y}}^2 = 4\mathrm{ax}, \\ \mathrm{then} \mathrm{we} \mathrm{know}, \mathrm{c} = \frac{\mathrm{a}}{{\mathrm{m}}_1} \\ {\mathrm{y}}^2=8\mathrm{x} \mathrm{and} \mathrm{y}=-\mathrm{x}+\mathrm{c} \\ 4\mathrm{a}=8 \\ \mathrm{a}=2 \\ \mathrm{c}=\frac{\mathrm{a}}{{\mathrm{m}}_1}=\frac{2}{-1}=-2 \\ \mathrm{hence} \mathrm{other} \mathrm{tangent} \mathrm{will} \mathrm{be} \mathrm{y}=-\mathrm{x}-2 \\ \mathrm{Point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{both} \mathrm{tangents}: \\ \mathrm{y}=\mathrm{x}+2 ....\left(1\right) \\ \mathrm{y}=-\mathrm{x}-2 ....\left(2\right) \\ \mathrm{from} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get} \\ \mathrm{x}+2=-\mathrm{x}-2 \\ 2\mathrm{x}=-4 \\ \mathrm{x}=-2 \\ \mathrm{y}=\mathrm{x}+2=-2+2=0 \\ \mathbf{hence}\mathbf{ }\mathbf{point}\mathbf{ }\mathbf{is}\mathbf{ }\left(\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{0}\right)\mathbf{.} \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards